最近我在面試中被要求破解一個(gè)程序，并且成功拿到了工作。
大家好，開(kāi)了一個(gè)新博客我真是非常激動(dòng)吶，于是我計(jì)劃每周都要更新幾篇文章。
看了標(biāo)題各位應(yīng)該知道本文的大概內(nèi)容了，這里我會(huì)講述一個(gè)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(qǐng)的是這樣一個(gè)職位，面試時(shí)候他們問(wèn)了我一些低級(jí)的問(wèn)題，有些我能答上來(lái)，還有些卻不行。
然后他們給我發(fā)了一封郵件，其中包含了一個(gè)加密的程序（CRACK MEEE!）。
回家之后我下載了它，打開(kāi)發(fā)現(xiàn)需要一個(gè)密碼來(lái)解鎖，心想原來(lái)他們是要我找到這個(gè)密碼。:)
乍看之下這是相當(dāng)困難的一件事，不過(guò)下面我會(huì)介紹破解過(guò)程中一些主要的概念。:)
第一件事就是在終端中執(zhí)行程序
root@lisa:~# ./CrackTheDoor
*** DOOR CONTROL SYSTEM ***
PASSWORD:
我隨手嘗試了幾個(gè)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(guān)于這個(gè)二進(jìn)制包更詳細(xì)的信息:)
接下來(lái)：
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(gè)命令。我稍微解釋一下，linux-gate.so這個(gè)文件應(yīng)該可以在你的電腦上找到。從ldd命令的結(jié)果可以看出它是一個(gè)共享庫(kù)對(duì)吧？那么你們聽(tīng)說(shuō)過(guò)Virtual DSO（Virtual Dynamic Shared Object）嗎？
這里建議閱讀一下關(guān)于的linux-gate.so詳細(xì)介紹。
libc.so.6是GNU系統(tǒng)上一個(gè)通用的C語(yǔ)言庫(kù)，這個(gè)你們可能都知道。
ld-linux.so是linux的動(dòng)態(tài)加載器。
到目前為止一切都還算順利，那么使用調(diào)試器來(lái)運(yùn)行這個(gè)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(jié)果程序自己就掛掉了，應(yīng)該是我們使用調(diào)試器造成的，看來(lái)有些反調(diào)試的小花招集成在里面了。那么……
我們來(lái)重新運(yùn)行它，并且找到程序是從哪開(kāi)始執(zhí)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(shè)置一個(gè)斷點(diǎn)再調(diào)試看看：
b * 0x804762c
按r運(yùn)行，應(yīng)該會(huì)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(jié)果應(yīng)該上面那樣的，這里語(yǔ)法系統(tǒng)是AT&A，你也可以切換為Intel語(yǔ)法體系。我個(gè)人而言，更偏向Intel語(yǔ)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?yàn)樗鼤?huì)中斷你的進(jìn)程，正如評(píng)論中所說(shuō)，這在x86架構(gòu)中是斷點(diǎn)指令（譯注：INT 3）。當(dāng)調(diào)試器想要中斷程序時(shí)，會(huì)將0xCC這個(gè)字節(jié)放到需要斷點(diǎn)的地方。
在0×8047666處有個(gè)指令repnz scas，意思是在段寄存器ES和偏移量EDI范圍的內(nèi)存中尋找AL里的值（0xCC）。
簡(jiǎn)單來(lái)說(shuō)，這行就是會(huì)搜尋內(nèi)存中的0xCC，如果找到就讓程序掛掉。
不過(guò)我不想在這花費(fèi)太多時(shí)間，我們來(lái)執(zhí)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(gè)函數(shù)的系統(tǒng)調(diào)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(gè)進(jìn)程，并且改變執(zhí)行狀態(tài)，就像調(diào)試器中一樣。
調(diào)試器經(jīng)常使用ptrace:)，畢竟這就是它們的任務(wù)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(shuō)一句，你只能使用一次ptrace[PTRACE_TRACEME]。因此如果調(diào)試器在這之前使用ptrace，那么我們的調(diào)用就會(huì)返回false，于是就能知道還有其他東西在控制程序了。
現(xiàn)在需要做的是繞過(guò)ptrace保護(hù)機(jī)制，使程序無(wú)法檢測(cè)到自己運(yùn)行在調(diào)試器中。
所以一個(gè)可能的策略是改變這個(gè)系統(tǒng)調(diào)用的返回值。
系統(tǒng)調(diào)用是用戶態(tài)與核心態(tài)之間的橋梁，于是可以確定ptrace中一定用到了一些系統(tǒng)調(diào)用來(lái)控制進(jìn)程。
我們希望檢測(cè)到程序使用ptrace，并且設(shè)置它的返回值為0:)
我在home目錄下建立了一個(gè)文件叫.gdbinit，這樣一來(lái)，當(dāng)運(yùn)行g(shù)db的時(shí)候，其中的配置信息就會(huì)被自動(dòng)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(huì)保存系統(tǒng)調(diào)用的返回值，這樣它就永遠(yuǎn)都是0了，或者說(shuō)是true。
通過(guò)這種方式，我們繞過(guò)了反調(diào)試機(jī)制，那就回到gdb繼續(xù)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(lái)調(diào)試了。:)
我在函數(shù)PJeGPC4TIVaKFmmy53DJ處又設(shè)置了一個(gè)斷點(diǎn)：
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(lè)子的步驟了。
我看到有些常量在內(nèi)存中移動(dòng)，并且和我輸入的值進(jìn)行了異或運(yùn)算。
繼續(xù)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(jié)果類似，現(xiàn)在我在其中增加一些常量：
下面是這個(gè)函數(shù)后面運(yùn)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(yùn)行的模式了嗎？如果沒(méi)看出來(lái)也沒(méi)關(guān)系……
在這個(gè)函數(shù)中，程序同樣將我的輸入和一些常量進(jìn)行了異或運(yùn)算。
現(xiàn)在來(lái)看看輸入，首先輸入會(huì)和這些常量進(jìn)行異或，然后將結(jié)果和另一些常量進(jìn)行對(duì)比。
因此這兩個(gè)函數(shù)應(yīng)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(gè)簡(jiǎn)單的python腳本將上面兩個(gè)常量進(jìn)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(gè)公司又給我發(fā)來(lái)了第二輪的破解題目，也一樣很有趣，下次我會(huì)在文章中寫到的。
譯注：原文評(píng)論中較多的疑問(wèn)是如何定位到函數(shù)PJeGPC4TIVaKFmmy53DJ，有人給出了合理的解釋，作者也默認(rèn)了：其實(shí)很簡(jiǎn)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。