最近我在面試中被要求破解一個程序，并且成功拿到了工作。
大家好，開了一個新博客我真是非常激動吶，于是我計劃每周都要更新幾篇文章。
看了標題各位應該知道本文的大概內容了，這里我會講述一個在土耳其的安卡拉面試的故事。
“軟件安全工程師（Software Security Engineer）”——我申請的是這樣一個職位，面試時候他們問了我一些低級的問題，有些我能答上來，還有些卻不行。
然后他們給我發了一封郵件，其中包含了一個加密的程序（CRACK MEEE!）。
回家之后我下載了它，打開發現需要一個密碼來解鎖，心想原來他們是要我找到這個密碼。:)
乍看之下這是相當困難的一件事，不過下面我會介紹破解過程中一些主要的概念。:)
第一件事就是在終端中執行程序
root@lisa:~# ./CrackTheDoor
*** DOOR CONTROL SYSTEM ***
PASSWORD:
我隨手嘗試了幾個愚蠢的密碼，3次之后，它就退出了。:)
再嘗試用工具去分析它，比如獲取文件信息：
root
@lisa:~# file CrackTheDoor
CrackTheDoor: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV),
dynamically linked (uses shared libs), for GNU/Linux 2.6.15,
BuildID[sha1]=0x9927be2fe310bea01d412164103b9c8b2d7567ea, not stripped
root
@lisa:~#
于是我們就得到了關于這個二進制包更詳細的信息:)
接下來：
root
@lisa:~# ldd CrackTheDoor
linux-gate.so.1 => (0xf777b000)
libc.so.6 => /lib32/libc.so.6 (0xf760c000)
/lib/ld-linux.so.2 (0xf777c000)
root
@lisa:~#
哎呀，順手就打了這個命令。我稍微解釋一下，linux-gate.so這個文件應該可以在你的電腦上找到。從ldd命令的結果可以看出它是一個共享庫對吧？那么你們聽說過Virtual DSO（Virtual Dynamic Shared Object）嗎？
這里建議閱讀一下關于的linux-gate.so詳細介紹。
libc.so.6是GNU系統上一個通用的C語言庫，這個你們可能都知道。
ld-linux.so是linux的動態加載器。
到目前為止一切都還算順利，那么使用調試器來運行這個程序試試：
root
@lisa:~# gdb CrackTheDoor
GNU gdb (GDB) 7.4.1-debian
Copyright (C) 2012 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law. Type “show copying”
and “show warranty” for details.
This GDB was configured as “x86_64-linux-gnu”.
For bug reporting instructions, please see:
<http://www.gnu.org/software/gdb/bugs/>…
Reading symbols from /root/CrackTheDoor…(no debugging symbols found)…done.
(gdb) r
Starting program: /root/CrackTheDoorProgram received signal SIGSEGV, Segmentation fault.
0x080484fb in __do_global_dtors_aux ()
(gdb)
結果程序自己就掛掉了，應該是我們使用調試器造成的，看來有些反調試的小花招集成在里面了。那么……
我們來重新運行它，并且找到程序是從哪開始執行的：
root
@lisa:~# gdb CrackTheDoor
GNU gdb (GDB) 7.4.1-debian
Copyright (C) 2012 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law. Type “show copying”
and “show warranty” for details.
This GDB was configured as “x86_64-linux-gnu”.
For bug reporting instructions, please see:
<http://www.gnu.org/software/gdb/bugs/>…
Reading symbols from /root/CrackTheDoor…(no debugging symbols found)…done.
(gdb) info file
Symbols from “/root/CrackTheDoor”.
Local exec file:
`/root/CrackTheDoor’, file type elf32-i386.
Entry point: 0x804762c
…
…
這樣就得到了程序的入口，在這里設置一個斷點再調試看看：
b * 0x804762c
按r運行，應該會停在入口的第一行：
gdb) x/30i $pc
=> 0x804762c: pusha
0x804762d: mov $0xaa,%dl
0x804762f: mov $0×8048480,%edi
0×8047634: mov $0x8048cbc,%ecx
0×8047639: mov %edi,0x80476f3
0x804763f: mov %ecx,0x80476f7
0×8047645: sub %edi,%ecx
0×8047647: mov $0x804762f,%esi
0x804764c: push $0x80476c1
0×8047651: pusha
0×8047652: mov $0×55,%al
0×8047654: xor $0×99,%al
0×8047656: mov $0×8047656,%edi
0x804765b: mov $0x80476e5,%ecx
0×8047660: sub $0×8047656,%ecx
0×8047666: repnz scas %es:(%edi),%al
0×8047668: je 0x804770a
0x804766e: mov %edi,0x80476eb
0×8047674: popa
0×8047675: add 0x80476eb,%edx
0x804767b: ret
結果應該上面那樣的，這里語法系統是AT&A，你也可以切換為Intel語法體系。我個人而言，更偏向Intel語法。
在地址0×8047654處，我們首先吧0×55放到AL寄存器，然后將它和0×99異或，得到了0xCC。
0xCC非常重要，因為它會中斷你的進程，正如評論中所說，這在x86架構中是斷點指令（譯注：INT 3）。當調試器想要中斷程序時，會將0xCC這個字節放到需要斷點的地方。
在0×8047666處有個指令repnz scas，意思是在段寄存器ES和偏移量EDI范圍的內存中尋找AL里的值（0xCC）。
簡單來說，這行就是會搜尋內存中的0xCC，如果找到就讓程序掛掉。
不過我不想在這花費太多時間，我們來執行strace命令：
root
@lisa:~# strace ./CrackTheDoor
execve(“./CrackTheDoor”, ["./CrackTheDoor"], [/* 17 vars */]) = 0
[ Process PID=31085 runs in 32 bit mode. ]
brk(0) = 0×9972000
access(“/etc/ld.so.nohwcap”, F_OK) = -1 ENOENT (No such file or directory)
mmap2(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0xfffffffff7715000
access(“/etc/ld.so.preload”, R_OK) = -1 ENOENT (No such file or directory)
open(“/etc/ld.so.cache”, O_RDONLY) = 3
fstat64(3, {st_mode=S_IFREG|0644, st_size=35597, …}) = 0
mmap2(NULL, 35597, PROT_READ, MAP_PRIVATE, 3, 0) = 0xfffffffff770c000
close(3) = 0
access(“/etc/ld.so.nohwcap”, F_OK) = -1 ENOENT (No such file or directory)
open(“/lib32/libc.so.6″, O_RDONLY) = 3
read(3, “\177ELF\1\1\1\3\3\1\300o\1004″…, 512) = 512
fstat64(3, {st_mode=S_IFREG|0755, st_size=1441884, …}) = 0
mmap2(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0xfffffffff770b000
mmap2(NULL, 1456504, PROT_READ|PROT_EXEC, MAP_PRIVATE|MAP_DENYWRITE, 3, 0) = 0xfffffffff75a7000
mprotect(0xf7704000, 4096, PROT_NONE) = 0
mmap2(0xf7705000, 12288, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_DENYWRITE, 3, 0x15d) = 0xfffffffff7705000
mmap2(0xf7708000, 10616, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_ANONYMOUS, -1, 0) = 0xfffffffff7708000
close(3) = 0
mmap2(NULL, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANONYMOUS, -1, 0) = 0xfffffffff75a6000
set_thread_area(0xffe4d864) = 0
mprotect(0xf7705000, 8192, PROT_READ) = 0
mprotect(0×8049000, 4096, PROT_READ) = 0
mprotect(0xf7733000, 4096, PROT_READ) = 0
munmap(0xf770c000, 35597) = 0
ptrace(PTRACE_TRACEME, 0, 0×1, 0) = -1 EPERM (Operation not permitted)
ptrace(PTRACE_TRACEME, 0, 0×1, 0) = -1 EPERM (Operation not permitted)
看到最后一行了吧，我們的程序又掛掉了，是ptrace這個函數的系統調用引起的。
在Linux中，ptrace是”Process Trace”的縮寫。有了它，你就可以控制另一個進程，并且改變執行狀態，就像調試器中一樣。
調試器經常使用ptrace:)，畢竟這就是它們的任務嘛。
我們可以想象一下這段代碼，可能是下面這樣的：
1 2 3 4 5 6 7 8 9 | int main()
{
if (ptrace(PTRACE_TRACEME, 0, 1, 0) < 0) {
printf("DEBUGGING... Bye\n");
return 1;
}
printf("Hello\n");
return 0;
}
|
順便說一句，你只能使用一次ptrace[PTRACE_TRACEME]。因此如果調試器在這之前使用ptrace，那么我們的調用就會返回false，于是就能知道還有其他東西在控制程序了。
現在需要做的是繞過ptrace保護機制，使程序無法檢測到自己運行在調試器中。
所以一個可能的策略是改變這個系統調用的返回值。
系統調用是用戶態與核心態之間的橋梁，于是可以確定ptrace中一定用到了一些系統調用來控制進程。
我們希望檢測到程序使用ptrace，并且設置它的返回值為0:)
我在home目錄下建立了一個文件叫.gdbinit，這樣一來，當運行gdb的時候，其中的配置信息就會被自動加載。
~/.gdbinit
set disassembly-flavor intel # Intel syntax is better
set disassemble-next-line on
catch syscall ptrace #Catch the syscall.
commands 1
set ($eax) = 0
continue
end
寄存器EAX會保存系統調用的返回值，這樣它就永遠都是0了，或者說是true。
通過這種方式，我們繞過了反調試機制，那就回到gdb繼續吧：
eren
@lisa:~$ gdb ./CrackTheDoor
GNU gdb (GDB) 7.4.1-debian
Copyright (C) 2012 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law. Type “show copying”
and “show warranty” for details.
This GDB was configured as “x86_64-linux-gnu”.
For bug reporting instructions, please see:
<http://www.gnu.org/software/gdb/bugs/>…
Catchpoint 1 (syscall ‘ptrace’ [26])
Reading symbols from /home/eren/CrackTheDoor…(no debugging symbols found)…done.
(gdb) r
Starting program: /home/eren/CrackTheDoorCatchpoint 1 (call to syscall ptrace), 0×08047698 in ?? ()
=> 0×08047698: 3d 00 f0 ff ff cmp eax,0xfffff000Catchpoint 1 (returned from syscall ptrace), 0×08047698 in ?? ()
=> 0×08047698: 3d 00 f0 ff ff cmp eax,0xfffff000
*** DOOR CONTROL SYSTEM ***
PASSWORD:
好了，至少目前為止我們能夠隨心所欲地來調試了。:)
我在函數PJeGPC4TIVaKFmmy53DJ處又設置了一個斷點：
Breakpoint 2, 0×08048534 in PJeGPC4TIVaKFmmy53DJ ()
=> 0×08048534 <PJeGPC4TIVaKFmmy53DJ+0>: 1e push ds
(gdb) x/40i $pc
=> 0×8048534 <PJeGPC4TIVaKFmmy53DJ>: push ds
0×8048535 <PJeGPC4TIVaKFmmy53DJ+1>: mov ebp,esp
0×8048537 <PJeGPC4TIVaKFmmy53DJ+3>: sub esp,0×20
0x804853a <PJeGPC4TIVaKFmmy53DJ+6>: mov BYTE PTR [ebp-0x1],0xe4
0x804853e <PJeGPC4TIVaKFmmy53DJ+10>: mov BYTE PTR [ebp-0x2],0×87
0×8048542 <PJeGPC4TIVaKFmmy53DJ+14>: mov BYTE PTR [ebp-0x3],0xfb
0×8048546 <PJeGPC4TIVaKFmmy53DJ+18>: mov BYTE PTR [ebp-0x4],0xbe
0x804854a <PJeGPC4TIVaKFmmy53DJ+22>: mov BYTE PTR [ebp-0x5],0xc9
0x804854e <PJeGPC4TIVaKFmmy53DJ+26>: mov BYTE PTR [ebp-0x6],0×93
0×8048552 <PJeGPC4TIVaKFmmy53DJ+30>: mov BYTE PTR [ebp-0x7],0×84
0×8048556 <PJeGPC4TIVaKFmmy53DJ+34>: mov BYTE PTR [ebp-0x8],0xfc
0x804855a <PJeGPC4TIVaKFmmy53DJ+38>: mov BYTE PTR [ebp-0x9],0x8d
0x804855e <PJeGPC4TIVaKFmmy53DJ+42>: mov BYTE PTR [ebp-0xa],0xe5
0×8048562 <PJeGPC4TIVaKFmmy53DJ+46>: mov BYTE PTR [ebp-0xb],0xbf
0×8048566 <PJeGPC4TIVaKFmmy53DJ+50>: mov BYTE PTR [ebp-0xc],0x5c
0x804856a <PJeGPC4TIVaKFmmy53DJ+54>: mov BYTE PTR [ebp-0xd],0xe2
0x804856e <PJeGPC4TIVaKFmmy53DJ+58>: mov BYTE PTR [ebp-0xe],0×76
0×8048572 <PJeGPC4TIVaKFmmy53DJ+62>: mov BYTE PTR [ebp-0xf],0×21
0×8048576 <PJeGPC4TIVaKFmmy53DJ+66>: mov BYTE PTR [ebp-0x10],0xb8
0x804857a <PJeGPC4TIVaKFmmy53DJ+70>: mov DWORD PTR [ebp-0x14],0×0
0×8048581 <PJeGPC4TIVaKFmmy53DJ+77>: mov eax,DWORD PTR [ebp-0x14]
0×8048584 <PJeGPC4TIVaKFmmy53DJ+80>: add eax,DWORD PTR [ebp+0x8]
0×8048587 <PJeGPC4TIVaKFmmy53DJ+83>: movzx eax,BYTE PTR [eax]
0x804858a <PJeGPC4TIVaKFmmy53DJ+86>: test al,al
0x804858c <PJeGPC4TIVaKFmmy53DJ+88>: je 0×8048808 <PJeGPC4TIVaKFmmy53DJ+724>
0×8048592 <PJeGPC4TIVaKFmmy53DJ+94>: mov eax,DWORD PTR [ebp-0x14]
0×8048595 <PJeGPC4TIVaKFmmy53DJ+97>: add eax,DWORD PTR [ebp+0x8]
0×8048598 <PJeGPC4TIVaKFmmy53DJ+100>: mov edx,DWORD PTR [ebp-0x14]
0x804859b <PJeGPC4TIVaKFmmy53DJ+103>: add edx,DWORD PTR [ebp+0x8]
0x804859e <PJeGPC4TIVaKFmmy53DJ+106>: movzx edx,BYTE PTR [edx]
0x80485a1 <PJeGPC4TIVaKFmmy53DJ+109>: xor dl,BYTE PTR [ebp-0x1]
0x80485a4 <PJeGPC4TIVaKFmmy53DJ+112>: mov BYTE PTR [eax],dl
0x80485a6 <PJeGPC4TIVaKFmmy53DJ+114>: add DWORD PTR [ebp-0x14],0×1
0x80485aa <PJeGPC4TIVaKFmmy53DJ+118>: mov eax,DWORD PTR [ebp-0x14]
0x80485ad <PJeGPC4TIVaKFmmy53DJ+121>: add eax,DWORD PTR [ebp+0x8]
0x80485b0 <PJeGPC4TIVaKFmmy53DJ+124>: movzx eax,BYTE PTR [eax]
0x80485b3 <PJeGPC4TIVaKFmmy53DJ+127>: test al,al
0x80485b5 <PJeGPC4TIVaKFmmy53DJ+129>: je 0x804880b <PJeGPC4TIVaKFmmy53DJ+727>
0x80485bb <PJeGPC4TIVaKFmmy53DJ+135>: mov eax,DWORD PTR [ebp-0x14]
0x80485be <PJeGPC4TIVaKFmmy53DJ+138>: add eax,DWORD PTR [ebp+0x8]
0x80485c1 <PJeGPC4TIVaKFmmy53DJ+141>: mov edx,DWORD PTR [ebp-0x14]
0x80485c4 <PJeGPC4TIVaKFmmy53DJ+144>: add edx,DWORD PTR [ebp+0x8]
0x80485c7 <PJeGPC4TIVaKFmmy53DJ+147>: movzx edx,BYTE PTR [edx]
0x80485ca <PJeGPC4TIVaKFmmy53DJ+150>: xor dl,BYTE PTR [ebp-0x2]
終于到有樂子的步驟了。
我看到有些常量在內存中移動，并且和我輸入的值進行了異或運算。
繼續深入研究……
(gdb) x/30i X1bdrhN8Yk9NZ59Vb7P2
0×8048838 <X1bdrhN8Yk9NZ59Vb7P2>: sbb ecx,DWORD PTR [ecx+0x20ec83e5]
0x804883e <X1bdrhN8Yk9NZ59Vb7P2+6>: mov DWORD PTR [ebp-0x18],0×0
0×8048845 <X1bdrhN8Yk9NZ59Vb7P2+13>: mov BYTE PTR [ebp-0x1],0xd9
0×8048849 <X1bdrhN8Yk9NZ59Vb7P2+17>: mov BYTE PTR [ebp-0x2],0xcd
0x804884d <X1bdrhN8Yk9NZ59Vb7P2+21>: mov BYTE PTR [ebp-0x3],0xc9
0×8048851 <X1bdrhN8Yk9NZ59Vb7P2+25>: mov BYTE PTR [ebp-0x4],0xe5
0×8048855 <X1bdrhN8Yk9NZ59Vb7P2+29>: mov BYTE PTR [ebp-0x5],0x9e
0×8048859 <X1bdrhN8Yk9NZ59Vb7P2+33>: mov BYTE PTR [ebp-0x6],0xd0
0x804885d <X1bdrhN8Yk9NZ59Vb7P2+37>: mov BYTE PTR [ebp-0x7],0xe8
0×8048861 <X1bdrhN8Yk9NZ59Vb7P2+41>: mov BYTE PTR [ebp-0x8],0xa5
0×8048865 <X1bdrhN8Yk9NZ59Vb7P2+45>: mov BYTE PTR [ebp-0x9],0xaf
0×8048869 <X1bdrhN8Yk9NZ59Vb7P2+49>: mov BYTE PTR [ebp-0xa],0×87
0x804886d <X1bdrhN8Yk9NZ59Vb7P2+53>: mov BYTE PTR [ebp-0xb],0xd2
0×8048871 <X1bdrhN8Yk9NZ59Vb7P2+57>: mov BYTE PTR [ebp-0xc],0×79
0×8048875 <X1bdrhN8Yk9NZ59Vb7P2+61>: mov BYTE PTR [ebp-0xd],0xa9
0×8048879 <X1bdrhN8Yk9NZ59Vb7P2+65>: mov BYTE PTR [ebp-0xe],0x5d
0x804887d <X1bdrhN8Yk9NZ59Vb7P2+69>: mov BYTE PTR [ebp-0xf],0×7
0×8048881 <X1bdrhN8Yk9NZ59Vb7P2+73>: mov BYTE PTR [ebp-0x10],0×81
0×8048885 <X1bdrhN8Yk9NZ59Vb7P2+77>: mov DWORD PTR [ebp-0x14],0×0
0x804888c <X1bdrhN8Yk9NZ59Vb7P2+84>: mov eax,DWORD PTR [ebp-0x14]
0x804888f <X1bdrhN8Yk9NZ59Vb7P2+87>: add eax,DWORD PTR [ebp+0x8]
0×8048892 <X1bdrhN8Yk9NZ59Vb7P2+90>: movzx eax,BYTE PTR [eax]
0×8048895 <X1bdrhN8Yk9NZ59Vb7P2+93>: cmp al,BYTE PTR [ebp-0x1]
0×8048898 <X1bdrhN8Yk9NZ59Vb7P2+96>: je 0x80488a2 <X1bdrhN8Yk9NZ59Vb7P2+106>
0x804889a <X1bdrhN8Yk9NZ59Vb7P2+98>: mov eax,DWORD PTR [ebp-0x18]
和之前得到的結果類似，現在我在其中增加一些常量：
下面是這個函數后面運行的情況：
0x804889d <X1bdrhN8Yk9NZ59Vb7P2+101>: jmp 0x8048a20 <X1bdrhN8Yk9NZ59Vb7P2+488>
0x80488a2 <X1bdrhN8Yk9NZ59Vb7P2+106>: add DWORD PTR [ebp-0x14],0×1
0x80488a6 <X1bdrhN8Yk9NZ59Vb7P2+110>: mov eax,DWORD PTR [ebp-0x14]
0x80488a9 <X1bdrhN8Yk9NZ59Vb7P2+113>: add eax,DWORD PTR [ebp+0x8]
0x80488ac <X1bdrhN8Yk9NZ59Vb7P2+116>: movzx eax,BYTE PTR [eax]
0x80488af <X1bdrhN8Yk9NZ59Vb7P2+119>: cmp al,BYTE PTR [ebp-0x2]
0x80488b2 <X1bdrhN8Yk9NZ59Vb7P2+122>: je 0x80488bc <X1bdrhN8Yk9NZ59Vb7P2+132>
0x80488b4 <X1bdrhN8Yk9NZ59Vb7P2+124>: mov eax,DWORD PTR [ebp-0x18]
0x80488b7 <X1bdrhN8Yk9NZ59Vb7P2+127>: jmp 0x8048a20 <X1bdrhN8Yk9NZ59Vb7P2+488>
0x80488bc <X1bdrhN8Yk9NZ59Vb7P2+132>: add DWORD PTR [ebp-0x14],0×1
0x80488c0 <X1bdrhN8Yk9NZ59Vb7P2+136>: mov eax,DWORD PTR [ebp-0x14]
0x80488c3 <X1bdrhN8Yk9NZ59Vb7P2+139>: add eax,DWORD PTR [ebp+0x8]
0x80488c6 <X1bdrhN8Yk9NZ59Vb7P2+142>: movzx eax,BYTE PTR [eax]
0x80488c9 <X1bdrhN8Yk9NZ59Vb7P2+145>: cmp al,BYTE PTR [ebp-0x3]
0x80488cc <X1bdrhN8Yk9NZ59Vb7P2+148>: je 0x80488d6 <X1bdrhN8Yk9NZ59Vb7P2+158>
0x80488ce <X1bdrhN8Yk9NZ59Vb7P2+150>: mov eax,DWORD PTR [ebp-0x18]
0x80488d1 <X1bdrhN8Yk9NZ59Vb7P2+153>: jmp 0x8048a20 <X1bdrhN8Yk9NZ59Vb7P2+488>
0x80488d6 <X1bdrhN8Yk9NZ59Vb7P2+158>: add DWORD PTR [ebp-0x14],0×1
0x80488da <X1bdrhN8Yk9NZ59Vb7P2+162>: mov eax,DWORD PTR [ebp-0x14]
0x80488dd <X1bdrhN8Yk9NZ59Vb7P2+165>: add eax,DWORD PTR [ebp+0x8]
—Type <return> to continue, or q <return> to quit—
0x80488e0 <X1bdrhN8Yk9NZ59Vb7P2+168>: movzx eax,BYTE PTR [eax]
0x80488e3 <X1bdrhN8Yk9NZ59Vb7P2+171>: cmp al,BYTE PTR [ebp-0x4]
0x80488e6 <X1bdrhN8Yk9NZ59Vb7P2+174>: je 0x80488f0 <X1bdrhN8Yk9NZ59Vb7P2+184>
0x80488e8 <X1bdrhN8Yk9NZ59Vb7P2+176>: mov eax,DWORD PTR [ebp-0x18]
0x80488eb <X1bdrhN8Yk9NZ59Vb7P2+179>: jmp 0x8048a20 <X1bdrhN8Yk9NZ59Vb7P2+488>
0x80488f0 <X1bdrhN8Yk9NZ59Vb7P2+184>: add DWORD PTR [ebp-0x14],0×1
0x80488f4 <X1bdrhN8Yk9NZ59Vb7P2+188>: mov eax,DWORD PTR [ebp-0x14]
0x80488f7 <X1bdrhN8Yk9NZ59Vb7P2+191>: add eax,DWORD PTR [ebp+0x8]
0x80488fa <X1bdrhN8Yk9NZ59Vb7P2+194>: movzx eax,BYTE PTR [eax]
0x80488fd <X1bdrhN8Yk9NZ59Vb7P2+197>: cmp al,BYTE PTR [ebp-0x5]
0×8048900 <X1bdrhN8Yk9NZ59Vb7P2+200>: je 0x804890a <X1bdrhN8Yk9NZ59Vb7P2+210>
0×8048902 <X1bdrhN8Yk9NZ59Vb7P2+202>: mov eax,DWORD PTR [ebp-0x18]
0×8048905 <X1bdrhN8Yk9NZ59Vb7P2+205>: jmp 0x8048a20 <X1bdrhN8Yk9NZ59Vb7P2+488>
0x804890a <X1bdrhN8Yk9NZ59Vb7P2+210>: add DWORD PTR [ebp-0x14],0×1
0x804890e <X1bdrhN8Yk9NZ59Vb7P2+214>: mov eax,DWORD PTR [ebp-0x14]
你們看出其中運行的模式了嗎？如果沒看出來也沒關系……
在這個函數中，程序同樣將我的輸入和一些常量進行了異或運算。
現在來看看輸入，首先輸入會和這些常量進行異或，然后將結果和另一些常量進行對比。
因此這兩個函數應該是這樣的：
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | void PJeGPC4TIVaKFmmy53DJ (int * p)
{
int array[] = {0xe4,0x87,0xfb,0xbe,0xc9,0x93,0x84,0xfc,
0x8d,0xe5,0xbf,0x5c,0xe2,0x76,0x21,0xb8}
for(i=0;i<16;i++)
{
p[i] = p[i] ^ array[i]
}
}
int X1bdrhN8Yk9NZ59Vb7P2(int * p)
{
int array = {0xd9,0xcd,0xc9,0xe5,0x9e,0xd0,0xe8,0xa5,
0xaf,0x87,0xd2,0x79,0xa9,0x5d,0x7,0x81}
for(i=0;i<16;i++)
{
if(p[i] != array[i])
return false; // fail..
}
return true
}
|
于是寫了一個簡單的python腳本將上面兩個常量進行異或操作，就得到密碼了：
#!/usr/bin/python
firstConst = [0xe4,0x87,0xfb,0xbe,0xc9,0x93,0x84,0xfc,0x8d,0xe5,0xbf,0x5c,0xe2,0x76,0x21,0xb8]
secondConst = [0xd9,0xcd,0xc9,0xe5,0x9e,0xd0,0xe8,0xa5,0xaf,0x87,0xd2,0x79,0xa9,0x5d,0x7,0x81]
ret =”"
for x in range(16):
ret+=chr(firstConst[x] ^ secondConst[x])
print ret
eren
@lisa:~$ ./CrackTheDoor*** DOOR CONTROL SYSTEM ***PASSWORD: =J2[WClY”bm%K+&9
*** ACCESS GRANTED ***
*** THE DOOR OPENED ***
看！成功了！
這個公司又給我發來了第二輪的破解題目，也一樣很有趣，下次我會在文章中寫到的。
譯注：原文評論中較多的疑問是如何定位到函數PJeGPC4TIVaKFmmy53DJ，有人給出了合理的解釋，作者也默認了：其實很簡單，不停地嘗試。也許寫下這篇文章之前有很多很多次失敗，但是讀者最感興趣的部分是成功的，因此那些都被省略了。