WINZIP的密码校对原理(3千字),WINZIP,加密算法 2008年06月23日 星期一 下午 12:45 WINZIP的密码校对用CRC,看输入密码经过计算同加密文件中的CRC是否相同,调用程序是WINZIP8.0的WZ32.DLL,过程如下: :20001E8A A1CC4D0620 mov eax, dword ptr [20064DCC] (密码) :20001E8F 85C0 test eax, eax :20001E91 7416 je 20001EA9 :20001E93 8D45F4 lea eax, dword ptr [ebp-0C] :20001E96 50 push eax :20001E97 E824000000 call 20001EC0 (CRC校对开始) . . . :20001EC0 55 push ebp :20001EC1 8BEC mov ebp, esp :20001EC3 83EC0C sub esp, 0000000C :20001EC6 A1CC4D0620 mov eax, dword ptr [20064DCC] :20001ECB 56 push esi :20001ECC 50 push eax :20001ECD E83EFDFFFF call 20001C10 (第一计算) :20001ED2 8B4D08 mov ecx, dword ptr [ebp 08] :20001ED5 83C404 add esp, 00000004 :20001ED8 33F6 xor esi, esi :20001EDA 8B11 mov edx, dword ptr [ecx] :20001EDC 8955F4 mov dword ptr [ebp-0C], edx :20001EDF 8B4104 mov eax, dword ptr [ecx 04] :20001EE2 8945F8 mov dword ptr [ebp-08], eax :20001EE5 8B4908 mov ecx, dword ptr [ecx 08] :20001EE8 894DFC mov dword ptr [ebp-04], ecx * Referenced by a (U)nconditional or (C)onditional Jump at Address: |:20001F0E(C) | :20001EEB E890FCFFFF call 20001B80 (第2计算) :20001EF0 8A5435F4 mov dl, byte ptr [ebp esi-0C] :20001EF4 32D0 xor dl, al :20001EF6 8AC2 mov al, dl :20001EF8 885435F4 mov byte ptr [ebp esi-0C], dl :20001EFC 25FF000000 and eax, 000000FF :20001F01 50 push eax :20001F02 E899FCFFFF call 20001BA0 (第3计算) :20001F07 83C404 add esp, 00000004 :20001F0A 46 inc esi :20001F0B 83FE0C cmp esi, 0000000C :20001F0E 7CDB jl 20001EEB (按密码长度反复计算) :20001F10 8B15240F0320 mov edx, dword ptr [20030F24] :20001F16 660FB645FF movzx ax, byte ptr [ebp-01] (计算结果) :20001F1B F6422002 test [edx 20], 02 :20001F1F 7414 je 20001F35 -------》JUMP * Referenced by a (U)nconditional or (C)onditional Jump at Address: |:20001F1F(C) | :20001F35 8B156A170820 mov edx, dword ptr [2008176A] (文件的CRC码) :20001F3B C1EA18 shr edx, 18 :20001F3E 663BC2 cmp ax, dx :20001F41 7407 je 20001F4A --相等就合法JUMP 文件的CRC码在WINZIP中查看属性,或文件的HEX地址:0000000E---00000011 关键问题:谁能写一个程序,按照文件的CRC码反计算密码,普度众生! KINGSUN 2001-08-29 版权所有 |
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