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[原创]新手分析个简单的CrackMe
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发表于:
2012-4-5 20:36
3628
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CrackMe技术等级自测系统 下载的级别2的…………
F12断点法……
00401570 |. 33C0 xor eax,eax
00401572 |. 33DB xor ebx,ebx
00401574 |. 33C9 xor ecx,ecx
00401576 |. B9 01000000 mov ecx,0x1
0040157B |. 33D2 xor edx,edx
0040157D |. 8B45 E4 mov eax,[local.7] eax=zxqloveyou
7A 78 71 6C 6F 76 65 79 6F 75
00401580 |> 8A18 /mov bl,byte ptr ds:[eax] 1 2 3 4 5 6 7 8 9 A
00401582 |. 32D9 |xor bl,cl 7B 7A 72 68 6A 70 62 71 66 7F
00401584 |. 8818 |mov byte ptr ds:[eax],bl
00401586 |. 41 |inc ecx
00401587 |. 40 |inc eax 指向下一位
00401588 |. 8038 00 |cmp byte ptr ds:[eax],0x0 判断用户名是否转换完成
0040158B |.^ 75 F3 \jnz short CrackMe.00401580
0040158D |. 33C0 xor eax,eax
0040158F |. 33DB xor ebx,ebx
00401591 |. 33C9 xor ecx,ecx
00401593 |. B9 0A000000 mov ecx,0xA
00401598 |. 33D2 xor edx,edx
0040159A |. 8B45 F0 mov eax,[local.4] eax=123456
31 32 33 34 35 36
0040159D |> 8A18 /mov bl,byte ptr ds:[eax] A B C D E F
0040159F |. 32D9 |xor bl,cl 3B 39 3F 39 3B 39
004015A1 |. 8818 |mov byte ptr ds:[eax],bl
004015A3 |. 41 |inc ecx
004015A4 |. 40 |inc eax
004015A5 |. 8038 00 |cmp byte ptr ds:[eax],0x0
004015A8 |.^ 75 F3 \jnz short CrackMe.0040159D
004015AA |. 8B45 E4 mov eax,[local.7] local.7=用户名
004015AD |. 8B55 F0 mov edx,[local.4] local.4=注册码
004015B0 |> 33C9 /xor ecx,ecx
004015B2 |. 8A18 |mov bl,byte ptr ds:[eax]
004015B4 |. 8A0A |mov cl,byte ptr ds:[edx]
004015B6 |. 3AD9 |cmp bl,cl
004015B8 75 09 jnz short CrackMe.004015C3 跳向不成功
004015BA |. 40 |inc eax
004015BB |. 42 |inc edx
004015BC |. 8038 00 |cmp byte ptr ds:[eax],0x0
004015BF |.^ 75 EF \jnz short CrackMe.004015B0 判断是否比较完成
004015C1 |. EB 16 jmp short CrackMe.004015D9 跳向成功
注册机源码:
#include<stdio.h>
#include <string.h>
#include <Windows.h>
main()
{
short int namelen,i=0;
char name[100]={0},serial[100]={0};
printf("Please input your name:");
scanf("%s",&name);
namelen=strlen(name);
for(;i<namelen;i++)
{
name[i] = name[i]^(i+1);
}
for(i=0;i<namelen;i++)
{
serial[i] = name[i]^((i+10));
}
printf("\nThis is your password:%s\n",serial);
system("pause");
return 0;
}
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