📊
CTF-All-In-One
  • 简介
  • 前言
  • 一、基础知识篇
    • 1.1 CTF 简介
    • 1.2 学习方法
    • 1.3 Linux 基础
    • 1.4 Web 安全基础
      • 1.4.1 HTML 基础
      • 1.4.2 HTTP 协议基础
      • 1.4.3 JavaScript 基础
      • 1.4.4 常见 Web 服务器基础
      • 1.4.5 OWASP Top Ten Project 漏洞基础
      • 1.4.6 PHP 源码审计基础
    • 1.5 逆向工程基础
      • 1.5.1 C/C++ 语言基础
      • 1.5.2 汇编基础
      • 1.5.3 Linux ELF
      • 1.5.4 Windows PE
      • 1.5.5 静态链接
      • 1.5.6 动态链接
      • 1.5.7 内存管理
      • 1.5.8 glibc malloc
      • 1.5.9 Linux 内核
      • 1.5.10 Windows 内核
      • 1.5.11 jemalloc
    • 1.6 密码学基础
      • 1.6.1 密码学导论
      • 1.6.2 流密码
      • 1.6.3 分组密码
      • 1.6.4 公钥密码
      • 1.6.5 消息认证和哈希函数
      • 1.6.6 数字签名
      • 1.6.7 密码协议
      • 1.6.8 密钥分配与管理
      • 1.6.9 数字货币
    • 1.7 Android 安全基础
      • 1.7.1 Android 环境搭建
      • 1.7.2 Dalvik 指令集
      • 1.7.3 ARM 汇编基础
      • 1.7.4 Android 常用工具
  • 二、工具篇
    • 虚拟化分析环境
      • 2.1.1 VirtualBox
      • 2.1.2 QEMU
      • 2.1.3 Docker
      • 2.1.4 Unicorn
    • 静态分析工具
      • 2.2.1 radare2
      • 2.2.2 IDA Pro
      • 2.2.3 JEB
      • 2.2.4 Capstone
      • 2.2.5 Keystone
      • 2.2.6 Ghidra
    • 动态分析工具
      • 2.3.1 GDB
      • 2.3.2 OllyDbg
      • 2.3.3 x64dbg
      • 2.3.4 WinDbg
      • 2.3.5 LLDB
    • 其他工具
      • 2.4.1 pwntools
      • 2.4.2 zio
      • 2.4.3 metasploit
      • 2.4.4 binwalk
      • 2.4.5 Burp Suite
      • 2.4.6 Wireshark
      • 2.4.7 Cuckoo Sandbox
  • 三、分类专题篇
    • Pwn
      • 3.1.1 格式化字符串漏洞
      • 3.1.2 整数溢出
      • 3.1.3 栈溢出
      • 3.1.4 返回导向编程(ROP)(x86)
      • 3.1.5 返回导向编程(ROP)(ARM)
      • 3.1.6 Linux 堆利用(一)
      • 3.1.7 Linux 堆利用(二)
      • 3.1.8 Linux 堆利用(三)
      • 3.1.9 Linux 堆利用(四)
      • 3.1.10 内核 ROP
      • 3.1.11 Linux 内核漏洞利用
      • 3.1.12 Windows 内核漏洞利用
      • 3.1.13 竞争条件
      • 3.1.14 虚拟机逃逸
    • Reverse
      • 3.2.1 patch 二进制文件
      • 3.2.2 脱壳技术(PE)
      • 3.2.3 脱壳技术(ELF)
      • 3.2.4 反调试技术(PE)
      • 3.2.5 反调试技术(ELF)
      • 3.2.6 指令混淆
    • Web
      • 3.3.1 SQL 注入利用
      • 3.3.2 XSS 漏洞利用
    • Crypto
    • Misc
      • 3.5.1 Lsb
    • Mobile
  • 四、技巧篇
    • 4.1 Linux 内核调试
    • 4.2 Linux 命令行技巧
    • 4.3 GCC 编译参数解析
    • 4.4 GCC 堆栈保护技术
    • 4.5 ROP 防御技术
    • 4.6 one-gadget RCE
    • 4.7 通用 gadget
    • 4.8 使用 DynELF 泄露函数地址
    • 4.9 shellcode 开发
    • 4.10 跳转导向编程(JOP)
    • 4.11 利用 mprotect 修改栈权限
    • 4.12 利用 __stack_chk_fail
    • 4.13 利用 _IO_FILE 结构
    • 4.14 glibc tcache 机制
    • 4.15 利用 vsyscall 和 vDSO
  • 五、高级篇
    • 5.0 软件漏洞分析
    • 5.1 模糊测试
      • 5.1.1 AFL fuzzer
      • 5.1.2 libFuzzer
    • 5.2 动态二进制插桩
      • 5.2.1 Pin
      • 5.2.2 DynamoRio
      • 5.2.3 Valgrind
    • 5.3 符号执行
      • 5.3.1 angr
      • 5.3.2 Triton
      • 5.3.3 KLEE
      • 5.3.4 S²E
    • 5.4 数据流分析
      • 5.4.1 Soot
    • 5.5 污点分析
      • 5.5.1 TaintCheck
    • 5.6 LLVM
      • 5.6.1 Clang
    • 5.7 程序切片
    • 5.8 SAT/SMT
      • 5.8.1 Z3
    • 5.9 基于模式的漏洞分析
    • 5.10 基于二进制比对的漏洞分析
    • 5.11 反编译技术
      • 5.11.1 RetDec
  • 六、题解篇
    • Pwn
      • 6.1.1 pwn HCTF2016 brop
      • 6.1.2 pwn NJCTF2017 pingme
      • 6.1.3 pwn XDCTF2015 pwn200
      • 6.1.4 pwn BackdoorCTF2017 Fun-Signals
      • 6.1.5 pwn GreHackCTF2017 beerfighter
      • 6.1.6 pwn DefconCTF2015 fuckup
      • 6.1.7 pwn 0CTF2015 freenote
      • 6.1.8 pwn DCTF2017 Flex
      • 6.1.9 pwn RHme3 Exploitation
      • 6.1.10 pwn 0CTF2017 BabyHeap2017
      • 6.1.11 pwn 9447CTF2015 Search-Engine
      • 6.1.12 pwn N1CTF2018 vote
      • 6.1.13 pwn 34C3CTF2017 readme_revenge
      • 6.1.14 pwn 32C3CTF2015 readme
      • 6.1.15 pwn 34C3CTF2017 SimpleGC
      • 6.1.16 pwn HITBCTF2017 1000levels
      • 6.1.17 pwn SECCONCTF2016 jmper
      • 6.1.18 pwn HITBCTF2017 Sentosa
      • 6.1.19 pwn HITBCTF2018 gundam
      • 6.1.20 pwn 33C3CTF2016 babyfengshui
      • 6.1.21 pwn HITCONCTF2016 Secret_Holder
      • 6.1.22 pwn HITCONCTF2016 Sleepy_Holder
      • 6.1.23 pwn BCTF2016 bcloud
      • 6.1.24 pwn HITCONCTF2016 House_of_Orange
      • 6.1.25 pwn HCTF2017 babyprintf
      • 6.1.26 pwn 34C3CTF2017 300
      • 6.1.27 pwn SECCONCTF2016 tinypad
      • 6.1.28 pwn ASISCTF2016 b00ks
      • 6.1.29 pwn Insomni'hack_teaserCTF2017 The_Great_Escape_part-3
      • 6.1.30 pwn HITCONCTF2017 Ghost_in_the_heap
      • 6.1.31 pwn HITBCTF2018 mutepig
      • 6.1.32 pwn SECCONCTF2017 vm_no_fun
      • 6.1.33 pwn 34C3CTF2017 LFA
      • 6.1.34 pwn N1CTF2018 memsafety
      • 6.1.35 pwn 0CTF2018 heapstorm2
      • 6.1.36 pwn NJCTF2017 messager
      • 6.1.37 pwn sixstarctf2018 babystack
      • 6.1.38 pwn HITCONCMT2017 pwn200
      • 6.1.39 pwn BCTF2018 house_of_Atum
      • 6.1.40 pwn LCTF2016 pwn200
      • 6.1.41 pwn PlaidCTF2015 PlaidDB
      • 6.1.42 pwn hacklu2015 bookstore
      • 6.1.43 pwn 0CTF2018 babyheap
      • 6.1.44 pwn ASIS2017 start_hard
      • 6.1.45 pwn LCTF2016 pwn100
    • Reverse
      • 6.2.1 re XHPCTF2017 dont_panic
      • 6.2.2 re ECTF2016 tayy
      • 6.2.3 re CodegateCTF2017 angrybird
      • 6.2.4 re CSAWCTF2015 wyvern
      • 6.2.5 re PicoCTF2014 Baleful
      • 6.2.6 re SECCONCTF2017 printf_machine
      • 6.2.7 re CodegateCTF2018 RedVelvet
      • 6.2.8 re DefcampCTF2015 entry_language
    • Web
      • 6.3.1 web HCTF2017 babycrack
    • Crypto
    • Misc
    • Mobile
  • 七、实战篇
    • CVE
      • 7.1.1 CVE-2017-11543 tcpdump sliplink_print 栈溢出漏洞
      • 7.1.2 CVE-2015-0235 glibc __nss_hostname_digits_dots 堆溢出漏洞
      • 7.1.3 CVE-2016-4971 wget 任意文件上传漏洞
      • 7.1.4 CVE-2017-13089 wget skip_short_body 栈溢出漏洞
      • 7.1.5 CVE–2018-1000001 glibc realpath 缓冲区下溢漏洞
      • 7.1.6 CVE-2017-9430 DNSTracer 栈溢出漏洞
      • 7.1.7 CVE-2018-6323 GNU binutils elf_object_p 整型溢出漏洞
      • 7.1.8 CVE-2010-2883 Adobe CoolType SING 表栈溢出漏洞
      • 7.1.9 CVE-2010-3333 Microsoft Word RTF pFragments 栈溢出漏洞
    • Malware
  • 八、学术篇
    • 8.1 The Geometry of Innocent Flesh on the Bone: Return-into-libc without Function Calls (on the x86)
    • 8.2 Return-Oriented Programming without Returns
    • 8.3 Return-Oriented Rootkits: Bypassing Kernel Code Integrity Protection Mechanisms
    • 8.4 ROPdefender: A Detection Tool to Defend Against Return-Oriented Programming Attacks
    • 8.5 Data-Oriented Programming: On the Expressiveness of Non-Control Data Attacks
    • 8.7 What Cannot Be Read, Cannot Be Leveraged? Revisiting Assumptions of JIT-ROP Defenses
    • 8.9 Symbolic Execution for Software Testing: Three Decades Later
    • 8.10 AEG: Automatic Exploit Generation
    • 8.11 Address Space Layout Permutation (ASLP): Towards Fine-Grained Randomization of Commodity Softwa
    • 8.13 New Frontiers of Reverse Engineering
    • 8.14 Who Allocated My Memory? Detecting Custom Memory Allocators in C Binaries
    • 8.21 Micro-Virtualization Memory Tracing to Detect and Prevent Spraying Attacks
    • 8.22 Practical Memory Checking With Dr. Memory
    • 8.23 Evaluating the Effectiveness of Current Anti-ROP Defenses
    • 8.24 How to Make ASLR Win the Clone Wars: Runtime Re-Randomization
    • 8.25 (State of) The Art of War: Offensive Techniques in Binary Analysis
    • 8.26 Driller: Augmenting Fuzzing Through Selective Symbolic Execution
    • 8.27 Firmalice - Automatic Detection of Authentication Bypass Vulnerabilities in Binary Firmware
    • 8.28 Cross-Architecture Bug Search in Binary Executables
    • 8.29 Dynamic Hooks: Hiding Control Flow Changes within Non-Control Data
    • 8.30 Preventing brute force attacks against stack canary protection on networking servers
    • 8.33 Under-Constrained Symbolic Execution: Correctness Checking for Real Code
    • 8.34 Enhancing Symbolic Execution with Veritesting
    • 8.38 TaintEraser: Protecting Sensitive Data Leaks Using Application-Level Taint Tracking
    • 8.39 DART: Directed Automated Random Testing
    • 8.40 EXE: Automatically Generating Inputs of Death
    • 8.41 IntPatch: Automatically Fix Integer-Overflow-to-Buffer-Overflow Vulnerability at Compile-Time
    • 8.42 Dynamic Taint Analysis for Automatic Detection, Analysis, and Signature Generation of Exploits
    • 8.43 DTA++: Dynamic Taint Analysis with Targeted Control-Flow Propagation
    • 8.44 Superset Disassembly: Statically Rewriting x86 Binaries Without Heuristics
    • 8.45 Ramblr: Making Reassembly Great Again
    • 8.46 FreeGuard: A Faster Secure Heap Allocator
    • 8.48 Reassembleable Disassembling
  • 九、附录
    • 9.1 更多 Linux 工具
    • 9.2 更多 Windows 工具
    • 9.3 更多资源
    • 9.4 Linux 系统调用表
    • 9.5 python2到3字符串转换
    • 9.6 幻灯片
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  • 题目解析
  • 逆向 VM 求解
  • 使用 Pin 求解
  • 参考资料

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  1. 六、题解篇
  2. Reverse

6.2.5 re PicoCTF2014 Baleful

Previous6.2.4 re CSAWCTF2015 wyvernNext6.2.6 re SECCONCTF2017 printf_machine

Last updated 3 years ago

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题目解析

$ file baleful
baleful: ELF 32-bit LSB executable, Intel 80386, version 1 (GNU/Linux), statically linked, stripped
$ strings baleful | grep -i upx
@UPX!
$Info: This file is packed with the UPX executable packer http://upx.sf.net $
$Id: UPX 3.91 Copyright (C) 1996-2013 the UPX Team. All Rights Reserved. $
UPX!u
UPX!
UPX!
$ upx -d baleful -o baleful_unpack
                       Ultimate Packer for eXecutables
                          Copyright (C) 1996 - 2017
UPX 3.94        Markus Oberhumer, Laszlo Molnar & John Reiser   May 12th 2017

        File size         Ratio      Format      Name
   --------------------   ------   -----------   -----------
    144956 <-      6752    4.66%   linux/i386    baleful_unpack

Unpacked 1 file.
$ file baleful_unpack
baleful_unpack: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked, interpreter /lib/ld-linux.so.2, for GNU/Linux 2.6.24, BuildID[sha1]=35d1a373cbe6a675ecbbc904722a86f853f20ce3, stripped

经过简单地检查,我们发现二进制文件被加了壳,使用 upx 脱掉就好了。

运行下看看,典型的密码验证题:

$ ./baleful_unpack
Please enter your password: ABCD
Sorry, wrong password!

逆向 VM 求解

打开 r2 开干吧!

[0x08048540]> pdf @ main
/ (fcn) main 96
|   main ();
|           ; var int local_8h @ ebp-0x8
|           ; var int local_10h @ esp+0x10
|           ; var int local_8ch @ esp+0x8c
|           ; DATA XREF from entry0 (0x8048557)
|           0x08049c82      push ebp
|           0x08049c83      mov ebp, esp
|           0x08049c85      push edi
|           0x08049c86      push ebx
|           0x08049c87      and esp, 0xfffffff0
|           0x08049c8a      sub esp, 0x90
|           0x08049c90      mov eax, dword gs:[0x14]                   ; [0x14:4]=-1 ; 20
|           0x08049c96      mov dword [local_8ch], eax
|           0x08049c9d      xor eax, eax
|           0x08049c9f      lea eax, [local_10h]                       ; 0x10 ; 16
|           0x08049ca3      mov ebx, eax
|           0x08049ca5      mov eax, 0
|           0x08049caa      mov edx, 0x1f                              ; 31
|           0x08049caf      mov edi, ebx
|           0x08049cb1      mov ecx, edx
|           0x08049cb3      rep stosd dword es:[edi], eax
|           0x08049cb5      lea eax, [local_10h]                       ; 0x10 ; 16
|           0x08049cb9      mov dword [esp], eax
|           0x08049cbc      call fcn.0804898b
|           0x08049cc1      mov eax, 0
|           0x08049cc6      mov edx, dword [local_8ch]                 ; [0x8c:4]=-1 ; 140
|           0x08049ccd      xor edx, dword gs:[0x14]
|       ,=< 0x08049cd4      je 0x8049cdb
|       |   0x08049cd6      call sym.imp.__stack_chk_fail              ; void __stack_chk_fail(void)
|       |   ; CODE XREF from main (0x8049cd4)
|       `-> 0x08049cdb      lea esp, [local_8h]
|           0x08049cde      pop ebx
|           0x08049cdf      pop edi
|           0x08049ce0      pop ebp
\           0x08049ce1      ret

fcn.0804898b 是程序主要的逻辑所在,很容易看出来它其实是实现了一个虚拟机:

使用 Pin 求解

就像上面那样逆向实在是太难了,不如 Pin 的黑科技。

编译 32 位 pintool:

[ManualExamples]$ make obj-ia32/inscount0.so TARGET=

随便输入几个长度不同的密码试试:

[ManualExamples]$ echo "A" | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack ; cat inscount.out
Please enter your password: Sorry, wrong password!
Count 437603
[ManualExamples]$ echo "AA" | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack ; cat inscount.out
Please enter your password: Sorry, wrong password!
Count 438397
[ManualExamples]$ echo "AAA" | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack ; cat inscount.out
Please enter your password: Sorry, wrong password!
Count 439191
$ python -c 'print(439191 - 438397)'
794
$ python -c 'print(438397 - 437603)'
794

指令执行的次数呈递增趋势,完美,这样只要递增到这个次数有不同时,就可以得到正确的密码长度:

#!/usr/bin/env python

import os

def get_count(flag):
    cmd = "echo " + "\"" + flag + "\"" + " | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack"
    os.system(cmd)
    with open("inscount.out") as f:
        count = int(f.read().split(" ")[1])
    return count

flag = "A"
p_count = get_count(flag)
for i in range(50):
    flag += "A"
    count = get_count(flag)
    print("count: ", count)
    diff = count - p_count
    print("diff: ", diff)
    if diff != 794:
        break
    p_count = count
print("length of password: ", len(flag))
Please enter your password: Sorry, wrong password!
count:  459041
diff:  794
Please enter your password: Sorry, wrong password!
count:  459835
diff:  794
Please enter your password: Sorry, wrong password!
count:  508273
diff:  48438
length of password:  30

好,密码长度为 30,接下来是逐字符爆破,首先要确定字符不同对 count 没有影响:

[ManualExamples]$ echo "A" | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack ; cat inscount.out
Please enter your password: Sorry, wrong password!
Count 437603
[ManualExamples]$ echo "b" | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack ; cat inscount.out
Please enter your password: Sorry, wrong password!
Count 437603
[ManualExamples]$ echo "_" | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack ; cat inscount.out
Please enter your password: Sorry, wrong password!
Count 437603

确实没有,写下脚本:

#!/usr/bin/env python

import os

def get_count(flag):
    cmd = "echo " + "\"" + flag + "\"" + " | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack"
    os.system(cmd)
    with open("inscount.out") as f:
        count = int(f.read().split(" ")[1])
    return count

charset = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_-+*'"

flag = list("A" * 30)
p_count = get_count("".join(flag))
for i in range(30):
    for c in charset:
        flag[i] = c
        print("".join(flag))
        count = get_count("".join(flag))
        print("count: ", count)
        if count != p_count:
            break
    p_count = count
print("password: ", "".join(flag))
packers_and_vms_and_xors_oh_mx
Please enter your password: Sorry, wrong password!
count:  507925
packers_and_vms_and_xors_oh_my
Please enter your password: Congratulations!
count:  505068
password:  packers_and_vms_and_xors_oh_my

简单到想哭。

参考资料

Pico CTF 2014 : Baleful
下载文件
题目解析
逆向 VM 求解
使用 Pin 求解
参考资料