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270 lines
14 KiB
Markdown
270 lines
14 KiB
Markdown
# 6.2.3 re CodegateCTF2017 angrybird
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- [题目解析](#题目解析)
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- [参考资料](#参考资料)
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[下载文件](../src/writeup/6.2.3_re_codegatectf2017_angrybird)
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## 题目解析
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看题目就知道，这是一个会让我们抓狂的程序，事实也确实如此。
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```
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$ file angrybird_org
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angrybird: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=089c3a14bcd7ffb08e94645cea46f1162b171445, stripped
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```
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```
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$ ./angrybird_org
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$
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```
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一运行就退出，应该是需要程序流上有问题。
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`main` 函数的开头有一些坑需要 patch，才能使程序正常运行，然后经过很多很多轮的运算和判断，可以看到 main 函数长达 18555 行：
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```
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[0x00400600]> pd 60 @ main
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/ (fcn) main 18555
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| main ();
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| : ; DATA XREF from 0x0040061d (entry0)
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| : 0x00400761 55 push rbp
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| : 0x00400762 4889e5 mov rbp, rsp
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| : 0x00400765 4883c480 add rsp, 0xffffffffffffff80
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| : 0x00400769 64488b042528. mov rax, qword fs:[0x28] ; [0x28:8]=-1 ; '(' ; 40
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| : 0x00400772 488945f8 mov qword [local_8h], rax
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| : 0x00400776 31c0 xor eax, eax ; 将 eax 置 0
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| : 0x00400778 83f800 cmp eax, 0 ; 比较 eax 和 0
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| `=< 0x0040077b 0f845ffeffff je sym.imp.exit ; eax == 0 时退出，所以需要将 je 换成 jne，或者把上一行的 0 换成 1
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| 0x00400781 48c745901860. mov qword [local_70h], reloc.strncmp_24 ; 0x606018
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| 0x00400789 48c745982060. mov qword [local_68h], reloc.puts_32 ; 0x606020
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| 0x00400791 48c745a02860. mov qword [local_60h], reloc.__stack_chk_fail_40 ; 0x606028
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| 0x00400799 48c745a83860. mov qword [local_58h], reloc.__libc_start_main_56 ; 0x606038
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| 0x004007a1 b800000000 mov eax, 0
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| 0x004007a6 e84bffffff call sub.you_should_return_21_not_1_:__6f6 ; 该函数中需要返回 21
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| 0x004007ab 89458c mov dword [local_74h], eax ; [local_74] = 21
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| 0x004007ae b800000000 mov eax, 0
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| 0x004007b3 e854ffffff call sub.stack_check_70c ; 栈检查函数，直接 nop 掉，或者进入函数修改逻辑
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| 0x004007b8 b800000000 mov eax, 0
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| 0x004007bd e868ffffff call sub.hello_72a
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| 0x004007c2 488b15a75820. mov rdx, qword [obj.stdin] ; [0x606070:8]=0 ; 从标准输入读入
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| 0x004007c9 8b4d8c mov ecx, dword [local_74h]
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| 0x004007cc 488d45b0 lea rax, [local_50h]
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| 0x004007d0 89ce mov esi, ecx ; esi = 21
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| 0x004007d2 4889c7 mov rdi, rax
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| 0x004007d5 e8f6fdffff call sym.imp.fgets ; char *fgets(char *s, int size, FILE *stream) ; patch 成功后就能调用 fgets
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| 0x004007da 0fb655b0 movzx edx, byte [local_50h] ; 读入的第一个字符
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| 0x004007de 0fb645b1 movzx eax, byte [local_4fh] ; 读入的第二个字符
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| 0x004007e2 31d0 xor eax, edx
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| 0x004007e4 8845d0 mov byte [local_30h], al
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| 0x004007e7 0fb645d0 movzx eax, byte [local_30h]
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| 0x004007eb 3c0f cmp al, 0xf ; 15 ; 对处理后的输入字符做判断
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| ,=< 0x004007ed 7f14 jg 0x400803 ; 若不满足条件，跳转失败，程序退出
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| | 0x004007ef bf94504000 mov edi, str.melong ; 0x405094 ; "melong"
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| | 0x004007f4 e897fdffff call sym.imp.puts ; int puts(const char *s)
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| | 0x004007f9 bf01000000 mov edi, 1
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| | 0x004007fe e8ddfdffff call sym.imp.exit ; void exit(int status)
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| | ; JMP XREF from 0x004007ed (main)
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| `-> 0x00400803 0fb655b0 movzx edx, byte [local_50h] ; 第二轮运算
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| 0x00400807 0fb645b1 movzx eax, byte [local_4fh]
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| 0x0040080b 21d0 and eax, edx
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| 0x0040080d 8845d0 mov byte [local_30h], al
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| 0x00400810 0fb645d0 movzx eax, byte [local_30h]
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| 0x00400814 3c50 cmp al, 0x50 ; 'P' ; 80
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| ,=< 0x00400816 7e14 jle 0x40082c
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| | 0x00400818 bf94504000 mov edi, str.melong ; 0x405094 ; "melong"
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| | 0x0040081d e86efdffff call sym.imp.puts ; int puts(const char *s)
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| | 0x00400822 bf01000000 mov edi, 1
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| | 0x00400827 e8b4fdffff call sym.imp.exit ; void exit(int status)
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| | ; JMP XREF from 0x00400816 (main)
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| `-> 0x0040082c c645d000 mov byte [local_30h], 0 ; 第三轮运算
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| 0x00400830 0fb645d0 movzx eax, byte [local_30h]
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| 0x00400834 3c01 cmp al, 1 ; 1
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| ,=< 0x00400836 7e14 jle 0x40084c
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| | 0x00400838 bf94504000 mov edi, str.melong ; 0x405094 ; "melong"
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| | 0x0040083d e84efdffff call sym.imp.puts ; int puts(const char *s)
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| | 0x00400842 bf01000000 mov edi, 1
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| | 0x00400847 e894fdffff call sym.imp.exit ; void exit(int status)
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| | ; JMP XREF from 0x00400836 (main)
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| `-> 0x0040084c 0fb655c2 movzx edx, byte [local_3eh] ; 第 n 轮运算
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| 0x00400850 0fb645b1 movzx eax, byte [local_4fh]
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| 0x00400854 21d0 and eax, edx
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| 0x00400856 8845d0 mov byte [local_30h], al
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| 0x00400859 0fb645d0 movzx eax, byte [local_30h]
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```
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第一处 patch，将指令 `je` 改成 `jne`：
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```
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[0x00400600]> s 0x0040077b
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[0x0040077b]> pd 1
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| `=< 0x0040077b 0f845ffeffff je sym.imp.exit
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[0x0040077b]> wx 0f85
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[0x0040077b]> pd 1
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| `=< 0x0040077b 0f855ffeffff jne sym.imp.exit
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```
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第二处 patch，函数 `sub.you_should_return_21_not_1_:__6f6`：
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```
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[0x0040077b]> pdf @ sub.you_should_return_21_not_1_:__6f6
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/ (fcn) sub.you_should_return_21_not_1_:__6f6 22
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| sub.you_should_return_21_not_1_:__6f6 ();
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| ; CALL XREF from 0x004007a6 (main)
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| 0x004006f6 55 push rbp
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| 0x004006f7 4889e5 mov rbp, rsp
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| 0x004006fa bf64504000 mov edi, str.you_should_return_21_not_1_:_ ; 0x405064 ; "you should return 21 not 1 :("
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| 0x004006ff e88cfeffff call sym.imp.puts ; int puts(const char *s)
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| 0x00400704 8b0556592000 mov eax, dword [0x00606060] ; [0x606060:4]=1 ; 修改 [0x606060:4] = 21 = 0x15
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| 0x0040070a 5d pop rbp
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\ 0x0040070b c3 ret
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[0x0040077b]> ?v 21
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0x15
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[0x0040077b]> s 0x00606060
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[0x00606060]> px 16
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- offset - 0 1 2 3 4 5 6 7 8 9 A B C D E F 0123456789ABCDEF
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0x00606060 0100 0000 0000 0000 0000 0000 0000 0000 ................
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[0x00606060]> wx 15
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[0x00606060]> px 16
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- offset - 0 1 2 3 4 5 6 7 8 9 A B C D E F 0123456789ABCDEF
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0x00606060 1500 0000 0000 0000 0000 0000 0000 0000 ................
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```
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另外该函数结尾处指令是 `pop rbp`，而不是正确情况下的 `leave`，我们把它改过来：
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```
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[0x00606060]> s 0x0040070a
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[0x0040070a]> pd 1
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| 0x0040070a 5d pop rbp
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[0x0040070a]> wx c9
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[0x0040070a]> pd 1
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| 0x0040070a c9 leave
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```
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第三处 patch，将调用 `sub.stack_check_70c` 的指令直接 `nop` 掉：
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```
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[0x00606060]> pdf @ sub.stack_check_70c
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/ (fcn) sub.stack_check_70c 30
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| sub.stack_check_70c ();
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| : ; CALL XREF from 0x004007b3 (main)
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| : 0x0040070c 55 push rbp
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| : 0x0040070d 4889e5 mov rbp, rsp
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| : 0x00400710 bf82504000 mov edi, str.stack_check ; 0x405082 ; "stack check"
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| : 0x00400715 e876feffff call sym.imp.puts ; int puts(const char *s)
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| : 0x0040071a 678b0424 mov eax, dword [esp]
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| : 0x0040071e 83f800 cmp eax, 0
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| `=< 0x00400721 0f85b9feffff jne sym.imp.exit
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| 0x00400727 90 nop
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| 0x00400728 5d pop rbp
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\ 0x00400729 c3 ret
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[0x00400600]> s 0x004007b3
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[0x004007b3]> pd 1
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| 0x004007b3 e854ffffff call sub.stack_check_70c
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[0x004007b3]> wx 9090909090
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[0x004007b3]> pd 5
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| 0x004007b3 90 nop
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| 0x004007b4 90 nop
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| 0x004007b5 90 nop
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| 0x004007b6 90 nop
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| 0x004007b7 90 nop
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```
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第四处 patch 是将 `sub.hello_72a` 函数中的 `je` 改成 `jne`：
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```
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[0x0040077b]> pdf @ sub.hello_72a
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/ (fcn) sub.hello_72a 55
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| sub.hello_72a ();
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| ; var int local_8h @ rbp-0x8
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| ; CALL XREF from 0x004007bd (main)
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| 0x0040072a 55 push rbp
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| 0x0040072b 4889e5 mov rbp, rsp
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| 0x0040072e 4883ec10 sub rsp, 0x10
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| 0x00400732 48c745f83860. mov qword [local_8h], reloc.__libc_start_main_56 ; 0x606038
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| 0x0040073a 488b45f8 mov rax, qword [local_8h]
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| 0x0040073e ba05000000 mov edx, 5
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| 0x00400743 be8e504000 mov esi, str.hello ; 0x40508e ; "hello"
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| 0x00400748 4889c7 mov rdi, rax
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| 0x0040074b e830feffff call sym.imp.strncmp ; int strncmp(const char *s1, const char *s2, size_t n) ; 如果相等则返回 0
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| 0x00400750 85c0 test eax, eax
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| ,=< 0x00400752 740a je 0x40075e ; 如果 eax 为 0，则跳转
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| | 0x00400754 bf01000000 mov edi, 1
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| | 0x00400759 e882feffff call sym.imp.exit ; void exit(int status)
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| | ; JMP XREF from 0x00400752 (sub.hello_72a)
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| `-> 0x0040075e 90 nop
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| 0x0040075f c9 leave
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\ 0x00400760 c3 ret
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```
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总的来说就是修改了下面几个地方：
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```
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$ radiff2 angrybird_org angrybird_mod
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0x0000070a 5d => c9 0x0000070a
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0x00000722 85 => 84 0x00000722
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0x00000752 74 => 75 0x00000752
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0x0000077c 84 => 85 0x0000077c
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0x000007b3 e854ffffff => 9090909090 0x000007b3
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0x00006060 01 => 15 0x00006060
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```
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这样程序的运行就正常了，它从标准输入读入字符，进行一系列的判断，由于程序执行流非常长，我们不可能一个一个地去 patch。radare2 里输入命令 `VV @ main` 可以看到下面的东西：
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不如使用 angr 来解决它，指定好目标地址，让它运行到那儿，在大多数情况下，这种方法都是有效的。
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```
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[0x00400761]> pd -20 @ main+18555
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| 0x00404f8e d00f ror byte [rdi], 1
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| 0x00404f90 b645 mov dh, 0x45 ; 'E' ; 69
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| 0x00404f92 d03c78 sar byte [rax + rdi*2], 1
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| ,=< 0x00404f95 7e14 jle 0x404fab
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| | 0x00404f97 bf94504000 mov edi, str.melong ; 0x405094 ; "melong"
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| | 0x00404f9c e8efb5ffff call sym.imp.puts ; int puts(const char *s)
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| | 0x00404fa1 bf01000000 mov edi, 1
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| | 0x00404fa6 e835b6ffff call sym.imp.exit ; void exit(int status)
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| | ; JMP XREF from 0x00404f95 (main)
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| `-> 0x00404fab 488d45b0 lea rax, [local_50h]
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| 0x00404faf 4889c6 mov rsi, rax
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| 0x00404fb2 bf9b504000 mov edi, str.you_typed_:__s_n ; 0x40509b ; "you typed : %s\n"
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| 0x00404fb7 b800000000 mov eax, 0
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| 0x00404fbc e8efb5ffff call sym.imp.printf ; int printf(const char *format)
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| 0x00404fc1 b800000000 mov eax, 0
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| 0x00404fc6 488b4df8 mov rcx, qword [local_8h]
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| 0x00404fca 6448330c2528. xor rcx, qword fs:[0x28]
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| ,=< 0x00404fd3 7405 je 0x404fda
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| | 0x00404fd5 e8c6b5ffff call sym.imp.__stack_chk_fail ; void __stack_chk_fail(void)
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| | ; JMP XREF from 0x00404fd3 (main)
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| `-> 0x00404fda c9 leave ; 选择一个目标地址
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\ 0x00404fdb c3 ret
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```
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因为每次错误退出之前，都会调用 `puts` 函数，所以应该避免其出现，将地址设置为参数 avoid。
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```
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[0x00400600]> is~puts
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vaddr=0x00400590 paddr=0x00000590 ord=002 fwd=NONE sz=16 bind=GLOBAL type=FUNC name=imp.puts
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```
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对于使用 angr 来说，上面的 patch 完全没有必要，只要选择一个合适的初始化地址，如 `0x004007da`，也就是 `fget` 函数的下一条指令，就可以跑出结果：
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```python
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import angr
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main = 0x004007da
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find = 0x00404fda # leave;ret
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avoid = 0x00400590 # puts@plt
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p = angr.Project('./angrybird_org')
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init = p.factory.blank_state(addr=main)
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pg = p.factory.simgr(init, threads=4)
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ex = pg.explore(find=find, avoid=avoid)
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final = ex.found[0].state
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flag = final.posix.dumps(0)
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print "Flag:", final.posix.dumps(1)
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```
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Bingo!!!（不能保证每次都有效，多试几次）
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```
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$ python2 solve.py
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WARNING | 2017-12-03 17:33:58,544 | angr.state_plugins.symbolic_memory | Concretizing symbolic length. Much sad; think about implementing.
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Flag: you typed : Im_so_cute&pretty_:)<29>
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```
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然后用我们 patch 过的程序来验证 flag：
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```
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$ ./angrybird_mod
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you should return 21 not 1 :(
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Im_so_cute&pretty_:)
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you typed : Im_so_cute&pretty_:)
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```
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同样需要一定的运气才能通过，祝好运:)
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## 参考资料
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- https://ctftime.org/task/3375
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