[反汇编练习] 160个CrackMe之002。 本系列文章的目的是从一个没有任何经验的新手的角度(其实就是我自己),一步步尝试将160个CrackMe全部破解,如果可以,通过任何方式写出一个类似于注册机的东西。 其中,文章中按照如下逻辑编排(解决如下问题): 1、使用什么环境和工具 2、程序分析 3、思路分析和破解流程 4、注册机的探索
1、工具和环境: WinXP SP3 + 52Pojie六周年纪念版OD + PEID + 汇编金手指。 160个CrackMe的打包文件。 下载地址: http://pan.baidu.com/s/1xUWOY 密码: jbnq 注: 1、Win7系统对于模块和程序开启了随机初始地址的功能,会给分析带来很大的负担,所以不建议使用Win7进行分析。 2、以上工具都是在52PoJie论坛下的原版程序,NOD32不报毒,个人承诺绝对不会进行任何和木马病毒相关内容。
2、程序分析: 想要破解一个程序,必须先了解这个程序。所以,在破解过程中,对最初程序的分析很重要,他可以帮助我们理解作者的目的和意图,特别是对于注册码的处理细节,从而方便我们反向跟踪和推导。 和上一节一样,打开CHM,选择第二个Afkayas,保存下来。运行程序,崩溃,缺少msvbvm50.dll,程序无法运行。没办法,上网上搜索一个,放到程序同一级目录,再次运行,OK。程序界面如下: 这是一个标准的Name/Serial注册码方式,二话不说,来个伪码测试: Name: 111222 Serial: 3334444 点击OK,弹出了一个错误对话框,You Get Wrong,Try again! 3、思路分析和破解流程: 首先,按照经验,有对话框提示的程序可以通过堆栈查找调用的位置。方法如下:输入伪码,点击OK,弹出错误对话框,此时不要关闭这个对话框,切换到OD,点击暂停程序,然后Ctrl+K到堆栈视图,如下: 到了这里基本蒙圈了,竟然不是一般的C/C++代码,看看后面的Called from大部分来自msvbvm50,user32,根据名称百度一下,发现是使用VB的模块写的。再查看红色部分,发现表面上和对话框相关的只有msvbvm50.rtcmsgbox和user32.MessageBoxIndirectA,最主要的是msvbvm50.rtcMsgBox的调用来自AfKayas_.0040261C,所以我们基本可以断定rtcMsgBox是Vb中弹出对话框相关的东西,不要犹豫,选中它,右键->show call。 [Asm] 纯文本查看 复制代码 004025E3 . /EB 56 jmp short 0040263B
004025E5 > |68 C81B4000 push 00401BC8 ; UNICODE "You Get Wrong"
004025EA . |68 9C1B4000 push 00401B9C ; ASCII "\r"
004025EF . |FFD7 call edi
004025F1 . |8BD0 mov edx,eax
004025F3 . |8D4D E8 lea ecx,dword ptr ss:[ebp-0x18]
004025F6 . |FFD3 call ebx
004025F8 . |50 push eax
004025F9 . |68 E81B4000 push 00401BE8 ; UNICODE "Try Again"
004025FE . |FFD7 call edi
00402600 . |8945 CC mov dword ptr ss:[ebp-0x34],eax
00402603 . |8D45 94 lea eax,dword ptr ss:[ebp-0x6C]
00402606 . |8D4D A4 lea ecx,dword ptr ss:[ebp-0x5C]
00402609 . |50 push eax
0040260A . |8D55 B4 lea edx,dword ptr ss:[ebp-0x4C]
0040260D . |51 push ecx
0040260E . |52 push edx
0040260F . |8D45 C4 lea eax,dword ptr ss:[ebp-0x3C]
00402612 . |6A 00 push 0x0
00402614 . |50 push eax
00402615 . |C745 C4 08000>mov dword ptr ss:[ebp-0x3C],0x8
0040261C . |FF15 10414000 call dword ptr ds:[<&MSVBVM50.#595>] ; msvbvm50.rtcMsgBox
将代码稍微向上查看,发现最近的地方就有“Try again”和“You get wrong”的提示,我们基本就可以断定就是这里了。继续向上,看看判断的逻辑部分在哪里: [Asm] 纯文本查看 复制代码 00402569 . 83C4 0C add esp,0xC
0040256C . B9 04000280 mov ecx,0x80020004
00402571 . B8 0A000000 mov eax,0xA
00402576 . 894D 9C mov dword ptr ss:[ebp-0x64],ecx
00402579 . 66:85F6 test si,si
0040257C . 8945 94 mov dword ptr ss:[ebp-0x6C],eax
0040257F . 894D AC mov dword ptr ss:[ebp-0x54],ecx
00402582 . 8945 A4 mov dword ptr ss:[ebp-0x5C],eax
00402585 . 894D BC mov dword ptr ss:[ebp-0x44],ecx
00402588 . 8945 B4 mov dword ptr ss:[ebp-0x4C],eax
0040258B . 74 58 je short 004025E5 ; 判断应该是这里吧?
0040258D . 68 801B4000 push 00401B80 ; UNICODE "You Get It"
00402592 . 68 9C1B4000 push 00401B9C ; ASCII "\r"
00402597 . FFD7 call edi
00402599 . 8BD0 mov edx,eax
0040259B . 8D4D E8 lea ecx,dword ptr ss:[ebp-0x18]
0040259E . FFD3 call ebx
004025A0 . 50 push eax
004025A1 . 68 A81B4000 push 00401BA8 ; UNICODE "KeyGen It Now"
004025A6 . FFD7 call edi
004025A8 . 8D4D 94 lea ecx,dword ptr ss:[ebp-0x6C]
004025AB . 8945 CC mov dword ptr ss:[ebp-0x34],eax
004025AE . 8D55 A4 lea edx,dword ptr ss:[ebp-0x5C]
004025B1 . 51 push ecx
004025B2 . 8D45 B4 lea eax,dword ptr ss:[ebp-0x4C]
004025B5 . 52 push edx
004025B6 . 50 push eax
004025B7 . 8D4D C4 lea ecx,dword ptr ss:[ebp-0x3C]
004025BA . 6A 00 push 0x0
004025BC . 51 push ecx
004025BD . C745 C4 08000>mov dword ptr ss:[ebp-0x3C],0x8
004025C4 . FF15 10414000 call dword ptr ds:[<&MSVBVM50.#595>] ; msvbvm50.rtcMsgBox
004025CA . 8D4D E8 lea ecx,dword ptr ss:[ebp-0x18]
004025CD . FF15 80414000 call dword ptr ds:[<&MSVBVM50.__vbaFreeS>; msvbvm50.__vbaFreeStr
004025D3 . 8D55 94 lea edx,dword ptr ss:[ebp-0x6C]
004025D6 . 8D45 A4 lea eax,dword ptr ss:[ebp-0x5C]
004025D9 . 52 push edx
004025DA . 8D4D B4 lea ecx,dword ptr ss:[ebp-0x4C]
004025DD . 50 push eax
004025DE . 8D55 C4 lea edx,dword ptr ss:[ebp-0x3C]
004025E1 . 51 push ecx
004025E2 . 52 push edx
004025E3 . EB 56 jmp short 0040263B
004025E5 > 68 C81B4000 push 00401BC8 ; UNICODE "You Get Wrong"
004025EA . 68 9C1B4000 push 00401B9C ; ASCII "\r"
004025EF . FFD7 call edi
004025F1 . 8BD0 mov edx,eax
004025F3 . 8D4D E8 lea ecx,dword ptr ss:[ebp-0x18]
004025F6 . FFD3 call ebx
004025F8 . 50 push eax
004025F9 . 68 E81B4000 push 00401BE8 ; UNICODE "Try Again"
004025FE . FFD7 call edi
很简单吧?直接在”You get it ”的旁边就有一个JE跳转,在OD查看发现如果跳转实现就提示错误,不实现就提示正确,爆破开始啦!选中je那一行,右键->Binary->Fill with NOPs. 这时在程序里随意输入试试,哈哈,爆破成功!
4、注册机探索: 由于我们在跳转附近没有发现任何类注册码的东西,所以,我们需要将这里块的内容F8跟踪一下,大概地看看那些地方可能和注册码相关。我们向上查找,在这段程序的开头(即找到最近的那个retn,下面开头一般是push ebp等)下断点: 00402310 > \55 push ebp // 程序段头 单步F8跟踪,将重要信息添加注释,特别是和Name/Serial相关的东西,分析后代码如下: [Asm] 纯文本查看 复制代码 00402403 . FF15 04414000 call dword ptr ds:[<&MSVBVM50.__vbaHresultChec>; MSVBVM50.__vbaHresultCheckObj
00402409 > 8B95 50FFFFFF mov edx,dword ptr ss:[ebp-0xB0]
0040240F . 8B45 E4 mov eax,dword ptr ss:[ebp-0x1C]
00402412 . 50 push eax ; //eax=111222,name
00402413 . 8B1A mov ebx,dword ptr ds:[edx]
00402415 . FF15 E4404000 call dword ptr ds:[<&MSVBVM50.__vbaLenBstr>] ; MSVBVM50.__vbaLenBstr
0040241B . 8BF8 mov edi,eax ; edi=6
0040241D . 8B4D E8 mov ecx,dword ptr ss:[ebp-0x18] ; ecx=1111222地址
00402420 . 69FF FB7C0100 imul edi,edi,0x17CFB ; // 乘法,edi*0x17CFB
00402426 . 51 push ecx
00402427 . 0F80 91020000 jo 004026BE
0040242D . FF15 F8404000 call dword ptr ds:[<&MSVBVM50.#516>] ; MSVBVM50.rtcAnsiValueBstr
00402433 . 0FBFD0 movsx edx,ax
00402436 . 03FA add edi,edx ; // edi=edi+edx(0x31)
00402438 . 0F80 80020000 jo 004026BE
0040243E . 57 push edi
0040243F . FF15 E0404000 call dword ptr ds:[<&MSVBVM50.__vbaStrI4>] ; MSVBVM50.__vbaStrI4
00402445 . 8BD0 mov edx,eax ; // eax=585235
00402447 . 8D4D E0 lea ecx,dword ptr ss:[ebp-0x20]
0040244A . FF15 70414000 call dword ptr ds:[<&MSVBVM50.__vbaStrMove>] ; MSVBVM50.__vbaStrMove
00402450 . 8BBD 50FFFFFF mov edi,dword ptr ss:[ebp-0xB0]
00402456 . 50 push eax ; // 585235
00402457 . 57 push edi ; // 0091B51C
00402458 . FF93 A4000000 call dword ptr ds:[ebx+0xA4]
0040245E . 85C0 test eax,eax ; // ==0
00402460 . 7D 12 jge short 00402474
00402462 . 68 A4000000 push 0xA4
00402467 . 68 5C1B4000 push 00401B5C
0040246C . 57 push edi
0040246D . 50 push eax
0040246E . FF15 04414000 call dword ptr ds:[<&MSVBVM50.__vbaHresultChec>; MSVBVM50.__vbaHresultCheckObj
00402474 > 8D45 E0 lea eax,dword ptr ss:[ebp-0x20]
00402477 . 8D4D E4 lea ecx,dword ptr ss:[ebp-0x1C]
0040247A . 50 push eax
0040247B . 8D55 E8 lea edx,dword ptr ss:[ebp-0x18]
0040247E . 51 push ecx
0040247F . 52 push edx
00402480 . 6A 03 push 0x3
00402482 . FF15 5C414000 call dword ptr ds:[<&MSVBVM50.__vbaFreeStrList>; MSVBVM50.__vbaFreeStrList
00402488 . 83C4 10 add esp,0x10
0040248B . 8D45 D4 lea eax,dword ptr ss:[ebp-0x2C]
0040248E . 8D4D D8 lea ecx,dword ptr ss:[ebp-0x28]
00402491 . 8D55 DC lea edx,dword ptr ss:[ebp-0x24]
00402494 . 50 push eax
00402495 . 51 push ecx
00402496 . 52 push edx
00402497 . 6A 03 push 0x3
00402499 . FF15 F4404000 call dword ptr ds:[<&MSVBVM50.__vbaFreeObjList>; MSVBVM50.__vbaFreeObjList
0040249F . 8B06 mov eax,dword ptr ds:[esi]
004024A1 . 83C4 10 add esp,0x10
004024A4 . 56 push esi
004024A5 . FF90 04030000 call dword ptr ds:[eax+0x304]
004024AB . 8B1D 0C414000 mov ebx,dword ptr ds:[<&MSVBVM50.__vbaObjSet>] ; MSVBVM50.__vbaObjSet
004024B1 . 50 push eax
004024B2 . 8D45 DC lea eax,dword ptr ss:[ebp-0x24]
004024B5 . 50 push eax
004024B6 . FFD3 call ebx ; <&MSVBVM50.__vbaObjSet>
004024B8 . 8BF8 mov edi,eax
004024BA . 8D55 E8 lea edx,dword ptr ss:[ebp-0x18]
004024BD . 52 push edx
004024BE . 57 push edi
004024BF . 8B0F mov ecx,dword ptr ds:[edi]
004024C1 . FF91 A0000000 call dword ptr ds:[ecx+0xA0]
004024C7 . 85C0 test eax,eax ; eax=0,zf=1
004024C9 . 7D 12 jge short 004024DD
004024CB . 68 A0000000 push 0xA0
004024D0 . 68 5C1B4000 push 00401B5C
004024D5 . 57 push edi
004024D6 . 50 push eax
004024D7 . FF15 04414000 call dword ptr ds:[<&MSVBVM50.__vbaHresultChec>; MSVBVM50.__vbaHresultCheckObj
004024DD > 56 push esi
004024DE . FF95 40FFFFFF call dword ptr ss:[ebp-0xC0]
004024E4 . 50 push eax
004024E5 . 8D45 D8 lea eax,dword ptr ss:[ebp-0x28]
004024E8 . 50 push eax
004024E9 . FFD3 call ebx
004024EB . 8BF0 mov esi,eax
004024ED . 8D55 E4 lea edx,dword ptr ss:[ebp-0x1C]
004024F0 . 52 push edx
004024F1 . 56 push esi
004024F2 . 8B0E mov ecx,dword ptr ds:[esi]
004024F4 . FF91 A0000000 call dword ptr ds:[ecx+0xA0]
004024FA . 85C0 test eax,eax ; eax=0
004024FC . 7D 12 jge short 00402510
004024FE . 68 A0000000 push 0xA0
00402503 . 68 5C1B4000 push 00401B5C
00402508 . 56 push esi
00402509 . 50 push eax
0040250A . FF15 04414000 call dword ptr ds:[<&MSVBVM50.__vbaHresultChec>; MSVBVM50.__vbaHresultCheckObj
00402510 > 8B45 E8 mov eax,dword ptr ss:[ebp-0x18] ; eax=3334444
00402513 . 8B4D E4 mov ecx,dword ptr ss:[ebp-0x1C] ; ecx=585235
00402516 . 8B3D 00414000 mov edi,dword ptr ds:[<&MSVBVM50.__vbaStrCat>] ; MSVBVM50.__vbaStrCat
0040251C . 50 push eax ; eax=3334444
0040251D . 68 701B4000 push 00401B70 ; UNICODE "AKA-"
00402522 . 51 push ecx ; ecx=585235
00402523 . FFD7 call edi ; <&MSVBVM50.__vbaStrCat>
00402525 . 8B1D 70414000 mov ebx,dword ptr ds:[<&MSVBVM50.__vbaStrMove>>; MSVBVM50.__vbaStrMove
0040252B . 8BD0 mov edx,eax ; edx=eax=AKA-585235
0040252D . 8D4D E0 lea ecx,dword ptr ss:[ebp-0x20]
00402530 . FFD3 call ebx ; <&MSVBVM50.__vbaStrMove>
00402532 . 50 push eax
00402533 . FF15 28414000 call dword ptr ds:[<&MSVBVM50.__vbaStrCmp>] ; MSVBVM50.__vbaStrCmp
00402539 . 8BF0 mov esi,eax ; eax=-1
0040253B . 8D55 E0 lea edx,dword ptr ss:[ebp-0x20]
0040253E . F7DE neg esi ; 取补
00402540 . 8D45 E8 lea eax,dword ptr ss:[ebp-0x18]
00402543 . 52 push edx
00402544 . 1BF6 sbb esi,esi
00402546 . 8D4D E4 lea ecx,dword ptr ss:[ebp-0x1C]
00402549 . 50 push eax
0040254A . 46 inc esi
0040254B . 51 push ecx
0040254C . 6A 03 push 0x3
0040254E . F7DE neg esi
00402550 . FF15 5C414000 call dword ptr ds:[<&MSVBVM50.__vbaFreeStrList>; MSVBVM50.__vbaFreeStrList
00402556 . 83C4 10 add esp,0x10
00402559 . 8D55 D8 lea edx,dword ptr ss:[ebp-0x28]
0040255C . 8D45 DC lea eax,dword ptr ss:[ebp-0x24]
0040255F . 52 push edx
00402560 . 50 push eax
00402561 . 6A 02 push 0x2
00402563 . FF15 F4404000 call dword ptr ds:[<&MSVBVM50.__vbaFreeObjList>; MSVBVM50.__vbaFreeObjList
00402569 . 83C4 0C add esp,0xC
0040256C . B9 04000280 mov ecx,0x80020004
00402571 . B8 0A000000 mov eax,0xA
00402576 . 894D 9C mov dword ptr ss:[ebp-0x64],ecx
00402579 . 66:85F6 test si,si ; esi=0,ZF=1
0040257C . 8945 94 mov dword ptr ss:[ebp-0x6C],eax
0040257F . 894D AC mov dword ptr ss:[ebp-0x54],ecx
00402582 . 8945 A4 mov dword ptr ss:[ebp-0x5C],eax
00402585 . 894D BC mov dword ptr ss:[ebp-0x44],ecx
00402588 . 8945 B4 mov dword ptr ss:[ebp-0x4C],eax
0040258B 74 58 je short 004025E5 ; // 爆破关键跳,NOP
0040258D . 68 801B4000 push 00401B80 ; UNICODE "You Get It"
00402592 . 68 9C1B4000 push 00401B9C ; ASCII "\r"
00402597 . FFD7 call edi
其实代码很简单,经过一遍跟踪,基本都出来了,重点分析部分如下: [Asm] 纯文本查看 复制代码 00402412 . 50 push eax ; //eax=111222,name
00402413 . 8B1A mov ebx,dword ptr ds:[edx]
00402415 . FF15 E4404000 call dword ptr ds:[<&MSVBVM50.__vbaLenBstr>] ; MSVBVM50.__vbaLenBstr
0040241B . 8BF8 mov edi,eax ; edi=6
0040241D . 8B4D E8 mov ecx,dword ptr ss:[ebp-0x18] ; ecx=1111222地址
00402420 . 69FF FB7C0100 imul edi,edi,0x17CFB ; // 乘法,edi*0x17CFB
00402426 . 51 push ecx
00402427 . 0F80 91020000 jo 004026BE
0040242D . FF15 F8404000 call dword ptr ds:[<&MSVBVM50.#516>] ; MSVBVM50.rtcAnsiValueBstr
00402433 . 0FBFD0 movsx edx,ax
00402436 . 03FA add edi,edx ; // edi=edi+edx(0x31)
00402438 . 0F80 80020000 jo 004026BE
0040243E . 57 push edi
0040243F . FF15 E0404000 call dword ptr ds:[<&MSVBVM50.__vbaStrI4>] ; MSVBVM50.__vbaStrI4
00402445 . 8BD0 mov edx,eax ; // eax=585235
首先,eax为Name的地址,经过__vbaLenBstr计算的长度赋值给edi,然后edi=edi*0x17CFB,接着edi=edi+edx,而edx为Name字符串第一个字符Ansi的值,最后进过__vbaStrI4将edi的整数值转换为十进制字符串585235。 然后向下跟踪和字符串585235相关的东西: [Asm] 纯文本查看 复制代码 00402510 > 8B45 E8 mov eax,dword ptr ss:[ebp-0x18] ; eax=3334444
00402513 . 8B4D E4 mov ecx,dword ptr ss:[ebp-0x1C] ; ecx=585235
00402516 . 8B3D 00414000 mov edi,dword ptr ds:[<&MSVBVM50.__vbaStrCat>] ; MSVBVM50.__vbaStrCat
0040251C . 50 push eax ; eax=3334444
0040251D . 68 701B4000 push 00401B70 ; UNICODE "AKA-"
00402522 . 51 push ecx ; ecx=585235
00402523 . FFD7 call edi ; <&MSVBVM50.__vbaStrCat>
00402525 . 8B1D 70414000 mov ebx,dword ptr ds:[<&MSVBVM50.__vbaStrMove>>; MSVBVM50.__vbaStrMove
0040252B . 8BD0 mov edx,eax ; edx=eax=AKA-585235
0040252D . 8D4D E0 lea ecx,dword ptr ss:[ebp-0x20]
00402530 . FFD3 call ebx ; <&MSVBVM50.__vbaStrMove>
00402532 . 50 push eax
00402533 . FF15 28414000 call dword ptr ds:[<&MSVBVM50.__vbaStrCmp>] ; MSVBVM50.__vbaStrCmp
00402539 . 8BF0 mov esi,eax ; eax=-1
这段代码中将585235通过__vbaStrCat与”AKA-”连接到一起组成字符串”AKA-585235”,然后通过__vbaStrCmp函数与我们的Serial比较结果,最后返回值放到eax中,通过返回值确定是否正确。将这里的分析与上一块合到一起,很容易就可以得出最终的注册码。 小结一下:先取出注册码的长度len, 然后取出注册码第一个字符的ANSI值cName, 让后计算len*0x17CFB+cName,将计算的值转换为10进制文本,前面加上”AKA-”组成最后的注册码。 C/CPP代码如下: [C++] 纯文本查看 复制代码 // CrackMe160.cpp : 定义控制台应用程序的入口点。
// 002
#include "stdafx.h"
#include <stdio.h>
#include "iostream"
char buff[100] = {0};
int _tmain(int argc, _TCHAR* argv[])
{
printf("160CrackMe-002 Name/Serial\r\n\r\n");
printf("Name:");
gets_s(buff,100);
int nLen = strlen(buff);
if ( nLen > 0 )
{
int nRet = nLen * 0x17CFB;
nRet += buff[0];
printf("AKA-%d\r\n",nRet);
}else{
printf("Input error!\r\n");
}
system("pause");
return 0;
}
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