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223 lines
7.0 KiB
Markdown
223 lines
7.0 KiB
Markdown
# 6.2.5 re PicoCTF2014 Baleful
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- [题目解析](#题目解析)
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- [逆向 VM 求解](#逆向-vm-求解)
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- [使用 Pin 求解](#使用-pin-求解)
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- [参考资料](#参考资料)
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[下载文件](../src/writeup/6.2.5_re_picoctf2014_baleful)
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## 题目解析
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```text
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$ file baleful
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baleful: ELF 32-bit LSB executable, Intel 80386, version 1 (GNU/Linux), statically linked, stripped
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$ strings baleful | grep -i upx
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@UPX!
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$Info: This file is packed with the UPX executable packer http://upx.sf.net $
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$Id: UPX 3.91 Copyright (C) 1996-2013 the UPX Team. All Rights Reserved. $
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UPX!u
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UPX!
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UPX!
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$ upx -d baleful -o baleful_unpack
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Ultimate Packer for eXecutables
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Copyright (C) 1996 - 2017
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UPX 3.94 Markus Oberhumer, Laszlo Molnar & John Reiser May 12th 2017
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File size Ratio Format Name
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-------------------- ------ ----------- -----------
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144956 <- 6752 4.66% linux/i386 baleful_unpack
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Unpacked 1 file.
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$ file baleful_unpack
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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
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```
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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。
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运行下看看，典型的密码验证题：
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```text
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$ ./baleful_unpack
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Please enter your password: ABCD
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Sorry, wrong password!
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```
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### 逆向 VM 求解
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打开 r2 开干吧！
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```text
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[0x08048540]> pdf @ main
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/ (fcn) main 96
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| main ();
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| ; var int local_8h @ ebp-0x8
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| ; var int local_10h @ esp+0x10
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| ; var int local_8ch @ esp+0x8c
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| ; DATA XREF from entry0 (0x8048557)
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| 0x08049c82 push ebp
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| 0x08049c83 mov ebp, esp
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| 0x08049c85 push edi
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| 0x08049c86 push ebx
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| 0x08049c87 and esp, 0xfffffff0
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| 0x08049c8a sub esp, 0x90
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| 0x08049c90 mov eax, dword gs:[0x14] ; [0x14:4]=-1 ; 20
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| 0x08049c96 mov dword [local_8ch], eax
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| 0x08049c9d xor eax, eax
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| 0x08049c9f lea eax, [local_10h] ; 0x10 ; 16
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| 0x08049ca3 mov ebx, eax
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| 0x08049ca5 mov eax, 0
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| 0x08049caa mov edx, 0x1f ; 31
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| 0x08049caf mov edi, ebx
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| 0x08049cb1 mov ecx, edx
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| 0x08049cb3 rep stosd dword es:[edi], eax
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| 0x08049cb5 lea eax, [local_10h] ; 0x10 ; 16
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| 0x08049cb9 mov dword [esp], eax
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| 0x08049cbc call fcn.0804898b
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| 0x08049cc1 mov eax, 0
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| 0x08049cc6 mov edx, dword [local_8ch] ; [0x8c:4]=-1 ; 140
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| 0x08049ccd xor edx, dword gs:[0x14]
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| ,=< 0x08049cd4 je 0x8049cdb
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| | 0x08049cd6 call sym.imp.__stack_chk_fail ; void __stack_chk_fail(void)
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| | ; CODE XREF from main (0x8049cd4)
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| `-> 0x08049cdb lea esp, [local_8h]
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| 0x08049cde pop ebx
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| 0x08049cdf pop edi
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| 0x08049ce0 pop ebp
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\ 0x08049ce1 ret
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```
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`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：
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### 使用 Pin 求解
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就像上面那样逆向实在是太难了，不如 Pin 的黑科技。
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编译 32 位 pintool：
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```text
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[ManualExamples]$ make obj-ia32/inscount0.so TARGET=
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```
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随便输入几个长度不同的密码试试：
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```text
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[ManualExamples]$ echo "A" | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack ; cat inscount.out
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Please enter your password: Sorry, wrong password!
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Count 437603
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[ManualExamples]$ echo "AA" | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack ; cat inscount.out
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Please enter your password: Sorry, wrong password!
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Count 438397
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[ManualExamples]$ echo "AAA" | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack ; cat inscount.out
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Please enter your password: Sorry, wrong password!
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Count 439191
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```
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```text
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$ python -c 'print(439191 - 438397)'
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794
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$ python -c 'print(438397 - 437603)'
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794
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```
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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：
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```python
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#!/usr/bin/env python
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import os
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def get_count(flag):
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cmd = "echo " + "\"" + flag + "\"" + " | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack"
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os.system(cmd)
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with open("inscount.out") as f:
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count = int(f.read().split(" ")[1])
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return count
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flag = "A"
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p_count = get_count(flag)
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for i in range(50):
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flag += "A"
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count = get_count(flag)
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print("count: ", count)
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diff = count - p_count
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print("diff: ", diff)
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if diff != 794:
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break
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p_count = count
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print("length of password: ", len(flag))
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```
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```text
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Please enter your password: Sorry, wrong password!
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count: 459041
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diff: 794
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Please enter your password: Sorry, wrong password!
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count: 459835
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diff: 794
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Please enter your password: Sorry, wrong password!
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count: 508273
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diff: 48438
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length of password: 30
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```
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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：
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```text
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[ManualExamples]$ echo "A" | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack ; cat inscount.out
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Please enter your password: Sorry, wrong password!
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Count 437603
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[ManualExamples]$ echo "b" | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack ; cat inscount.out
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Please enter your password: Sorry, wrong password!
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Count 437603
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[ManualExamples]$ echo "_" | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack ; cat inscount.out
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Please enter your password: Sorry, wrong password!
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Count 437603
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```
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确实没有，写下脚本：
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```python
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#!/usr/bin/env python
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import os
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def get_count(flag):
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cmd = "echo " + "\"" + flag + "\"" + " | ../../../pin -t obj-ia32/inscount0.so -o inscount.out -- ~/baleful_unpack"
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os.system(cmd)
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with open("inscount.out") as f:
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count = int(f.read().split(" ")[1])
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return count
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charset = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789_-+*'"
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flag = list("A" * 30)
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p_count = get_count("".join(flag))
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for i in range(30):
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for c in charset:
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flag[i] = c
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print("".join(flag))
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count = get_count("".join(flag))
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print("count: ", count)
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if count != p_count:
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break
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p_count = count
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print("password: ", "".join(flag))
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```
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```text
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packers_and_vms_and_xors_oh_mx
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Please enter your password: Sorry, wrong password!
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count: 507925
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packers_and_vms_and_xors_oh_my
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Please enter your password: Congratulations!
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count: 505068
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password: packers_and_vms_and_xors_oh_my
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```
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简单到想哭。
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## 参考资料
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- [Pico CTF 2014 : Baleful](https://github.com/ctfs/write-ups-2014/tree/master/pico-ctf-2014/master-challenge/baleful-200)
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