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7.0 KiB
7.0 KiB
6.2.5 re PicoCTF2014 Baleful
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。