# 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)