上次那篇破文（不是破解文章，是破烂文章）丢了，我补在这贴的后面。这次说是中级，其实只难了一点点而已，cmp You,高手 jz offset NextPage。
好了，再来一篇。还是那个系列的，下载地址：
http://opencrackmes.crackmes.de/ope...ackmes/k4n2.zip
运行一下，呵呵，外观一模一样。反汇编，前面的部分几乎完全一样，GetDlgItem,GetWindowText，我都不写了，直接看下面，注意[ebp-2C]是用户名的长度。（可以先跳过去看后面的说明。）
代码:
:004010ED 837DD403 cmp dword ptr [ebp-2C], 00000003
:004010F1 0F8E38010000 jle 0040122F ;用户名必须大于3位
:004010F7 33D2 xor edx, edx
:004010F9 33DB xor ebx, ebx
:004010FB 8B55D4 mov edx, dword ptr [ebp-2C]
:004010FE 0155C4 add dword ptr [ebp-3C], edx
:00401101 0155C4 add dword ptr [ebp-3C], edx ;算出[EBP-3C]
:00401104 8BC2 mov eax, edx
:00401106 83C005 add eax, 00000005
:00401109 8945B8 mov dword ptr [ebp-48], eax ;算出[EBP-48]
:0040110C 33C0 xor eax, eax
:0040110E 8BCF mov ecx, edi
:00401110 83C104 add ecx, 00000004
:00401113 894DB4 mov dword ptr [ebp-4C], ecx ;算出[EBP-4C]
:00401116 33C9 xor ecx, ecx
:00401118 0155BC add dword ptr [ebp-44], edx
:0040111B 017DBC add dword ptr [ebp-44], edi ;算出[EBP-44]
:0040111E 6BFF03 imul edi, 00000003
:00401121 897DC0 mov dword ptr [ebp-40], edi ;算出[EBP-40]
:00401124 33FF xor edi, edi
:00401126 0FBE8C0544FFFFFF movsx ecx, byte ptr [ebp+eax-000000BC]
:0040112E 83F961 cmp ecx, 00000061
:00401131 7C07 jl 0040113A
:00401133 90 nop
:00401134 90 nop
:00401135 90 nop
:00401136 90 nop
:00401137 83E920 sub ecx, 00000020
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:00401131(C)
|
:0040113A 8BF1 mov esi, ecx
:0040113C 03DE add ebx, esi
:0040113E 0FAFD9 imul ebx, ecx
:00401141 4A dec edx
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:00401178(C)
|
:00401142 0FBE8C2F44FFFFFF movsx ecx, byte ptr [edi+ebp-000000BC]
:0040114A 0FBEB42F45FFFFFF movsx esi, byte ptr [edi+ebp-000000BB]
:00401152 83F961 cmp ecx, 00000061
:00401155 7D12 jge 00401169
:00401157 90 nop
:00401158 90 nop
:00401159 90 nop
:0040115A 90 nop
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:0040116C(U)
|
:0040115B 83FE61 cmp esi, 00000061
:0040115E 7D0E jge 0040116E
:00401160 90 nop
:00401161 90 nop
:00401162 90 nop
:00401163 90 nop
:00401164 EB0B jmp 00401171
:00401166 90 nop
:00401167 90 nop
:00401168 90 nop
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:00401155(C)
|
:00401169 83E920 sub ecx, 00000020
:0040116C EBED jmp 0040115B
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:0040115E(C)
|
:0040116E 83EE20 sub esi, 00000020
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:00401164(U)
|
:00401171 47 inc edi
:00401172 03DE add ebx, esi
:00401174 0FAFD9 imul ebx, ecx
:00401177 4A dec edx
:00401178 75C8 jne 00401142
:0040117A 895DC8 mov dword ptr [ebp-38], ebx ;算出[EBP-38]
:0040117D 33C9 xor ecx, ecx
:0040117F 33D2 xor edx, edx
:00401181 33DB xor ebx, ebx
:00401183 33C0 xor eax, eax
:00401185 837DD432 cmp dword ptr [ebp-2C], 00000032
:00401189 0F8DA0000000 jnl 0040122F
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:0040119F(C)
|
:0040118F 0FBE840D44FFFFFF movsx eax, byte ptr [ebp+ecx-000000BC]
:00401197 03C1 add eax, ecx
:00401199 03D8 add ebx, eax
:0040119B 41 inc ecx
:0040119C 3B4DD4 cmp ecx, dword ptr [ebp-2C]
:0040119F 75EE jne 0040118F
:004011A1 D1C0 rol eax, 1
:004011A3 3540E20100 xor eax, 0001E240
:004011A8 8945B0 mov dword ptr [ebp-50], eax ;算出[EBP-50]
:004011AB 33C9 xor ecx, ecx
:004011AD 33D2 xor edx, edx
:004011AF 33DB xor ebx, ebx
:004011B1 33C0 xor eax, eax
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:004011C6(C)
|
:004011B3 0FBE840D44FFFFFF movsx eax, byte ptr [ebp+ecx-000000BC]
:004011BB 6BD006 imul edx, eax, 00000006
:004011BE 33C2 xor eax, edx
:004011C0 03D8 add ebx, eax
:004011C2 41 inc ecx
:004011C3 3B4DD4 cmp ecx, dword ptr [ebp-2C]
:004011C6 75EB jne 004011B3
:004011C8 035DB0 add ebx, dword ptr [ebp-50]
:004011CB 895DAC mov dword ptr [ebp-54], ebx ;算出[EBP-54]
:004011CE FF75C0 push [ebp-40]
:004011D1 FF75C4 push [ebp-3C]
:004011D4 FF75BC push [ebp-44]
:004011D7 FF75C8 push [ebp-38]
:004011DA FF75B4 push [ebp-4C]
:004011DD FF75B8 push [ebp-48]
:004011E0 FF75AC push [ebp-54]
:004011E3 FF75B0 push [ebp-50]
* Possible StringData Ref from Data Obj ->"%lX%lu-%lu%lX-%lu%lu-%lX%lX"
|
:004011E6 6838B44000 push 0040B438
:004011EB 8D857CFEFFFF lea eax, dword ptr [ebp+FFFFFE7C]
:004011F1 50 push eax
:004011F2 E88D3D0000 call 00404F84 ;wsprinf()
:004011F7 83C428 add esp, 00000028
:004011FA 8D957CFEFFFF lea edx, dword ptr [ebp+FFFFFE7C]
:00401200 52 push edx
:00401201 8D8DE0FEFFFF lea ecx, dword ptr [ebp+FFFFFEE0]
:00401207 51 push ecx
* Reference To: KERNEL32.lstrcmpA, Ord:0000h
|
:00401208 E8399C0000 Call 0040AE46 ;比较
:0040120D 85C0 test eax, eax
:0040120F 750F jne 00401220 ;关键跳转
用我上篇文章介绍的方法找串式参考，然后向上找关键跳转。具体过程不说了，看看你是不是找的到。对了，就是40120F这个地方了。向上看看，有一个lstrcmp，上次已经说了，这个就是字符串比较。可以看到它的前面有两个PUSH作为比较的字符串，在这里下断点，看看两个字串是什么？D ecx，是我们输入的假注册码，D edx，是一个长长的字符串，当然就是真正的注册码啦。嗯，再向上看这个注册码是咋来的：
代码:
:004011CE FF75C0 push [ebp-40]
:004011D1 FF75C4 push [ebp-3C]
:004011D4 FF75BC push [ebp-44]
:004011D7 FF75C8 push [ebp-38]
:004011DA FF75B4 push [ebp-4C]
:004011DD FF75B8 push [ebp-48]
:004011E0 FF75AC push [ebp-54]
:004011E3 FF75B0 push [ebp-50]
* Possible StringData Ref from Data Obj ->"%lX%lu-%lu%lX-%lu%lu-%lX%lX"
|
:004011E6 6838B44000 push 0040B438
:004011EB 8D857CFEFFFF lea eax, dword ptr [ebp+FFFFFE7C]
:004011F1 50 push eax ;结果存在[ebp+FFFFFE7C]
:004011F2 E88D3D0000 call 00404F84 ;这个CALL其实是wsprinf
:004011F7 83C428 add esp, 00000028
呵呵，一个"%lX%lu-%lu%lX-%lu%lu-%lX%lX"。还记得上次的例子吗，那回是一个"%lX"，这回复杂了一些哟。别担心，还是很简单的。上次说了，"%lX"是十六进制的大写形式，那么"%lu"呢，就是普通的十进制形式啦。再看前面PUSH进了一堆参数，这些[ebp-xx]的形式都是函数里面的局部变量，在这里把它们以不同的形式表示出来再组合好，就是真正的注册码了。下一步的目标，当然就是看这8个变量是如何计算出的啦。记住这几个变量都是什么.从头看：
代码:
:004010FB 8B55D4 mov edx, dword ptr [ebp-2C] ;edx=[ebp-2C]是用户名的长度n
:004010FE 0155C4 add dword ptr [ebp-3C], edx ;[ebp-3C]可是一个重要变量,
:00401101 0155C4 add dword ptr [ebp-3C], edx ;[ebp-3C]=2n,看出来没有
:00401104 8BC2 mov eax, edx
:00401106 83C005 add eax, 00000005
:00401109 8945B8 mov dword ptr [ebp-48], eax ;[ebp-48]=n+5,也是一个重要变量
:0040110C 33C0 xor eax, eax
:0040110E 8BCF mov ecx, edi ;edi是一个常数64F4F0
:00401110 83C104 add ecx, 00000004
:00401113 894DB4 mov dword ptr [ebp-4C], ecx ;[ebp-4C]=64f4f4
:00401116 33C9 xor ecx, ecx
:00401118 0155BC add dword ptr [ebp-44], edx
:0040111B 017DBC add dword ptr [ebp-44], edi ;[ebp-44]=64f4f0+n
:0040111E 6BFF03 imul edi, 00000003
:00401121 897DC0 mov dword ptr [ebp-40], edi ;[ebp-40]=64f4f0*3
这几个计算都比较简单的，有5个变量已经被搞定了。其中那个EDI我实在没看明白和我们的输入有什么关系，我改动用户名和注册码它也不会变化，因此我认为这是一个常量，如果不对请高手指正。
代码:
:00401126 0FBE8C0544FFFFFF movsx ecx, byte ptr [ebp+eax-BC];[EBP-BC]是用户名,EAX作为指针
:0040112E 83F961 cmp ecx, 00000061
:00401131 7C07 jl 0040113A ;如果小于61即'a'就跳转
:00401137 83E920 sub ecx, 00000020 ;如果大于就减20,对于字母来说是小写转大写
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:00401131(C)
|
:0040113A 8BF1 mov esi, ecx ;ECX为用户名第一个字符
:0040113C 03DE add ebx, esi ;EBX=ECX
:0040113E 0FAFD9 imul ebx, ecx ;实际上EBX=ECX*ECX
:00401141 4A dec edx ;EDX为循环变量减1
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:00401178(C)
|
:00401142 0FBE8C2F44FFFFFF movsx ecx, byte ptr [edi+ebp-000000BC];前一个字符
:0040114A 0FBEB42F45FFFFFF movsx esi, byte ptr [edi+ebp-000000BB];后一个字符
;EDI也是控制取字符的指针,这里相当于每次取出两个字符,前一个放在ECX,后一个放在ESI
:00401152 83F961 cmp ecx, 00000061
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:0040116C(U)
|
:0040115B 83FE61 cmp esi, 00000061
:0040115E 7D0E jge 0040116E ;这里对字符进行同样的转换
:00401164 EB0B jmp 00401171
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:00401155(C)
|
:00401169 83E920 sub ecx, 00000020
:0040116C EBED jmp 0040115B
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:0040115E(C)
|
:0040116E 83EE20 sub esi, 00000020
:00401171 47 inc edi ;EDI这个指针+1
:00401172 03DE add ebx, esi ;EBX是累加的结果,再加上后一个字符
:00401174 0FAFD9 imul ebx, ecx ;再乘上前一个字符
:00401177 4A dec edx
:00401178 75C8 jne 00401142 ;是否取完?
:0040117A 895DC8 mov dword ptr [ebp-38], ebx ;累加结果存在[ebp-38]
这段麻烦一些,看不明白的话看我的注册机代码就清楚了.来看下面一段:
代码:
:0040118F 0FBE840D44FFFFFF movsx eax, byte ptr [ebp+ecx-000000BC];循环取字符
:00401197 03C1 add eax, ecx ;EAX=每位字符+ECX
:00401199 03D8 add ebx, eax ;累加到EBX
:0040119B 41 inc ecx ;循环变量递增
:0040119C 3B4DD4 cmp ecx, dword ptr [ebp-2C]
:0040119F 75EE jne 0040118F ;如果未取完则继续
:004011A1 D1C0 rol eax, 1 ;EAX左移1位
:004011A3 3540E20100 xor eax, 0001E240 ;EAX XOR 1E240
:004011A8 8945B0 mov dword ptr [ebp-50], eax
相信你已经对这个形式很熟悉了吧，[EBP-BC]这是用户名，然后用一个ECX循环递增来取每个字符，再看看：add eax,ecx / add ebx,eax 好像是把每位字符的值再和字符的位置（ECX）累加起来的，呵呵，作者开了个小玩笑。看看下面的操作，都是对EAX进行的，可是累加的结果是放在EBX中呀，其实EAX是用户名的最后一个字符加上用户的长度。要说下的是ROL，这个本来是“滚动”，但因为EAX肯定很小，最高位为0，所以在注册机中我简单的用左移SHL代替了。计算结果放在[ebp-50]。
代码:
:004011B3 0FBE840D44FFFFFF movsx eax, byte ptr [ebp+ecx-000000BC]
:004011BB 6BD006 imul edx, eax, 00000006
:004011BE 33C2 xor eax, edx
:004011C0 03D8 add ebx, eax
:004011C2 41 inc ecx
:004011C3 3B4DD4 cmp ecx, dword ptr [ebp-2C]
:004011C6 75EB jne 004011B3
:004011C8 035DB0 add ebx, dword ptr [ebp-50]
:004011CB 895DAC mov dword ptr [ebp-54], ebx
这一段不写说明了，当作测验，应该看懂了吧，算出结果放在[ebp-54]。
至此八个变量都出来了，注册机也容易了吧。
代码:
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
void main()
{
int len,i;
int EBP_40,EBP_3C,EBP_44,EBP_38,EBP_4C,EBP_48,EBP_54,EBP_50;
int EDI=0x64F4F0;
char name[50]={0};
printf("Please input your name:");
scanf("%s",name);
len=strlen(name);
EBP_3C=len*2;
EBP_48=len+5;
EBP_4C=EDI+4;
EBP_44=EDI+len;
EBP_40=EDI*3;
EBP_50=((len-1+name[len-1])<<1);
EBP_50^=0x1E240;
EBP_54=0;
for (i=0;i<len;i++)
EBP_54+=((name[i]*6)^name[i]);
EBP_54+=EBP_50;
for (i=0;i<len;i++)
if (name[i]>='a') name[i]-=0x20;
EBP_38=name[0]*name[0];
for (i=1;i<len;i++)
EBP_38=(EBP_38+name[i])*name[i-1];
printf("Your password is: %lX%lu-%lu%lX-%lu%lu-%lX%lX\n",
EBP_50,EBP_54,EBP_48,EBP_4C,EBP_38,EBP_44,EBP_3C,EBP_40);
printf("KeyGen by RoBa Enjoy Cracking!\n");
}
一个可用的注册码:
Name: RoBa
Serial: 1E288125744-964F4F4-29089574586616308-812EDED0
本文地址：http://com.8s8s.com/it/it26064.htm