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CTF-All-In-One/doc/6.2.5_re_picoctf2014_baleful.md
firmianay 7b743fc91e update 6.2.5
2017-12-06 17:19:49 +08:00

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# 6.2.5 re PicoCTF2014 Baleful
- [题目解析](#题目解析)
- [逆向 VM 求解](#逆向-vm-求解)
- [使用 Pin 求解](#使用-pin-求解)
- [参考资料](#参考资料)
[下载文件](../src/writeup/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 0x08048557 (entry0)
| 0x08049c82 55 push ebp
| 0x08049c83 89e5 mov ebp, esp
| 0x08049c85 57 push edi
| 0x08049c86 53 push ebx
| 0x08049c87 83e4f0 and esp, 0xfffffff0
| 0x08049c8a 81ec90000000 sub esp, 0x90
| 0x08049c90 65a114000000 mov eax, dword gs:[0x14] ; [0x14:4]=-1 ; 20
| 0x08049c96 8984248c0000. mov dword [local_8ch], eax
| 0x08049c9d 31c0 xor eax, eax
| 0x08049c9f 8d442410 lea eax, [local_10h] ; 0x10 ; 16
| 0x08049ca3 89c3 mov ebx, eax
| 0x08049ca5 b800000000 mov eax, 0
| 0x08049caa ba1f000000 mov edx, 0x1f ; 31
| 0x08049caf 89df mov edi, ebx
| 0x08049cb1 89d1 mov ecx, edx
| 0x08049cb3 f3ab rep stosd dword es:[edi], eax
| 0x08049cb5 8d442410 lea eax, [local_10h] ; 0x10 ; 16
| 0x08049cb9 890424 mov dword [esp], eax
| 0x08049cbc e8caecffff call fcn.0804898b
| 0x08049cc1 b800000000 mov eax, 0
| 0x08049cc6 8b94248c0000. mov edx, dword [local_8ch] ; [0x8c:4]=-1 ; 140
| 0x08049ccd 653315140000. xor edx, dword gs:[0x14]
| ,=< 0x08049cd4 7405 je 0x8049cdb
| | 0x08049cd6 e8e5e7ffff call sym.imp.__stack_chk_fail ; void __stack_chk_fail(void)
| | ; JMP XREF from 0x08049cd4 (main)
| `-> 0x08049cdb 8d65f8 lea esp, [local_8h]
| 0x08049cde 5b pop ebx
| 0x08049cdf 5f pop edi
| 0x08049ce0 5d pop ebp
\ 0x08049ce1 c3 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：
```
[0x08048540]> pdf @ fcn.0804898b
Linear size differs too much from the bbsum, please use pdr instead.
[0x08048540]> pdr @ fcn.0804898b
/ (fcn) fcn.0804898b 226
| fcn.0804898b (int arg_8h);
| ; var int local_b4h @ ebp-0xb4
| ; var int local_38h @ ebp-0x38
| ; var int local_34h @ ebp-0x34
| ; var int local_30h @ ebp-0x30
| ; var int local_2ch @ ebp-0x2c
| ; var int local_28h @ ebp-0x28
| ; var int local_24h @ ebp-0x24
| ; var int local_20h @ ebp-0x20
| ; var int local_1ch @ ebp-0x1c
| ; var int local_18h @ ebp-0x18
| ; var int local_14h @ ebp-0x14
| ; var int local_10h @ ebp-0x10
| ; var int local_ch @ ebp-0xc
| ; arg int arg_8h @ ebp+0x8
| ; CALL XREF from 0x08049cbc (main)
| 0x0804898b 55 push ebp
| 0x0804898c 89e5 mov ebp, esp
| 0x0804898e 81ecc8000000 sub esp, 0xc8
| 0x08048994 c745cc001000. mov dword [local_34h], 0x1000
| 0x0804899b 837d0800 cmp dword [arg_8h], 0
| 0x0804899f 742a je 0x80489cb
| ----------- true: 0x080489cb false: 0x080489a1
| 0x080489a1 c745d0000000. mov dword [local_30h], 0
| 0x080489a8 eb19 jmp 0x80489c3
| ----------- true: 0x080489c3
| ; JMP XREF from 0x080489c7 (fcn.0804898b)
| 0x080489aa 8b45d0 mov eax, dword [local_30h]
| 0x080489ad c1e002 shl eax, 2
| 0x080489b0 034508 add eax, dword [arg_8h]
| 0x080489b3 8b10 mov edx, dword [eax]
| 0x080489b5 8b45d0 mov eax, dword [local_30h]
| 0x080489b8 8994854cffff. mov dword [ebp + eax*4 - 0xb4], edx
| 0x080489bf 8345d001 add dword [local_30h], 1
| ----------- true: 0x080489c3
| ; JMP XREF from 0x080489a8 (fcn.0804898b)
| 0x080489c3 837dd01e cmp dword [local_30h], 0x1e ; [0x1e:4]=-1 ; 30
| 0x080489c7 7ee1 jle 0x80489aa
| ----------- true: 0x080489aa false: 0x080489c9
| 0x080489c9 eb21 jmp 0x80489ec
| ----------- true: 0x080489ec
| ; JMP XREF from 0x0804899f (fcn.0804898b)
| 0x080489cb c745d4000000. mov dword [local_2ch], 0
| 0x080489d2 eb12 jmp 0x80489e6
| ----------- true: 0x080489e6
| ; JMP XREF from 0x080489ea (fcn.0804898b)
| 0x080489d4 8b45d4 mov eax, dword [local_2ch]
| 0x080489d7 c784854cffff. mov dword [ebp + eax*4 - 0xb4], 0
| 0x080489e2 8345d401 add dword [local_2ch], 1
| ----------- true: 0x080489e6
| ; JMP XREF from 0x080489d2 (fcn.0804898b)
| 0x080489e6 837dd41e cmp dword [local_2ch], 0x1e ; [0x1e:4]=-1 ; 30
| 0x080489ea 7ee8 jle 0x80489d4
| ----------- true: 0x080489d4 false: 0x080489ec
| ; JMP XREF from 0x080489c9 (fcn.0804898b)
| 0x080489ec c745c800f000. mov dword [local_38h], 0xf000
| 0x080489f3 c745d8000000. mov dword [local_28h], 0
| 0x080489fa c745ec000000. mov dword [local_14h], 0
| 0x08048a01 c745f0000000. mov dword [local_10h], 0
| 0x08048a08 c745f4000000. mov dword [local_ch], 0
| 0x08048a0f c745e8000000. mov dword [local_18h], 0
| 0x08048a16 8b45e8 mov eax, dword [local_18h]
| 0x08048a19 8945e4 mov dword [local_1ch], eax
| 0x08048a1c 8b45e4 mov eax, dword [local_1ch]
| 0x08048a1f 8945e0 mov dword [local_20h], eax
| 0x08048a22 8b45e0 mov eax, dword [local_20h]
| 0x08048a25 8945dc mov dword [local_24h], eax
| 0x08048a28 e93a120000 jmp 0x8049c67
| ----------- true: 0x08049c67
| ; JMP XREF from 0x08049c74 (fcn.0804898b)
| 0x08048a2d 8b45cc mov eax, dword [local_34h]
| 0x08048a30 05c0c00408 add eax, 0x804c0c0
| 0x08048a35 0fb600 movzx eax, byte [eax]
| 0x08048a38 0fbec0 movsx eax, al
| 0x08048a3b 83f820 cmp eax, 0x20 ; 32
| 0x08048a3e 0f871e120000 ja 0x8049c62
| ----------- true: 0x08049c62 false: 0x08048a44
| 0x08048a44 8b0485d49d04. mov eax, dword [eax*4 + 0x8049dd4] ; [0x8049dd4:4]=0x8048a4d
| 0x08048a4b ffe0 jmp eax
| ; JMP XREF from 0x08048a3e (fcn.0804898b)
| 0x08049c62 8345cc01 add dword [local_34h], 1
| 0x08049c66 90 nop
| ----------- true: 0x08049c67
| ; XREFS: JMP 0x08048a28 JMP 0x08048a51 JMP 0x08048a8a JMP 0x08048bbf JMP 0x08048cf4 JMP 0x08048e2a JMP 0x08048f8c JMP 0x080490c1
| ; XREFS: JMP 0x080491f6 JMP 0x0804932b JMP 0x08049462 JMP 0x08049599 JMP 0x080495f0 JMP 0x08049644 JMP 0x08049698 JMP 0x080496cc
| ; XREFS: JMP 0x080496e7 JMP 0x08049710 JMP 0x08049739 JMP 0x08049762 JMP 0x0804978b JMP 0x080497b4 JMP 0x080497dd JMP 0x080498eb
| ; XREFS: JMP 0x080499fd JMP 0x08049a81 JMP 0x08049ab4 JMP 0x08049ae7 JMP 0x08049b3e JMP 0x08049b8d JMP 0x08049bf6 JMP 0x08049c2c
| ; XREFS: JMP 0x08049c60
| 0x08049c67 8b45cc mov eax, dword [local_34h]
| 0x08049c6a 05c0c00408 add eax, 0x804c0c0
| 0x08049c6f 0fb600 movzx eax, byte [eax]
| 0x08049c72 3c1d cmp al, 0x1d ; 29
| 0x08049c74 0f85b3edffff jne 0x8048a2d
| ----------- true: 0x08048a2d false: 0x08049c7a
| 0x08049c7a 8b854cffffff mov eax, dword [local_b4h]
| ; JMP XREF from 0x08048a6f (fcn.0804898b + 228)
| 0x08049c80 c9 leave
\ 0x08049c81 c3 ret
```
```
[0x08048540]> px @ 0x0804c060
- offset - 0 1 2 3 4 5 6 7 8 9 A B C D E F 0123456789ABCDEF
0x0804c060 7c86 0408 ad86 0408 d486 0408 a987 0408 |...............
0x0804c070 fb86 0408 1b87 0408 4e87 0408 d887 0408 ........N.......
0x0804c080 1986 0408 1388 0408 3488 0408 7b88 0408 ........4...{...
0x0804c090 b688 0408 f188 0408 2c89 0408 6086 0408 ........,...`...
0x0804c0a0 6a86 0408 6789 0408 0000 0000 0000 0000 j...g...........
```
#### 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：
```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：
```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](https://github.com/ctfs/write-ups-2014/tree/master/pico-ctf-2014/master-challenge/baleful-200)
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