diablo2002’s crackme1算法分析[5]

[入库:2005年8月19日] [更新:2007年3月24日]

本文简介:选择自 bmd2chen 的 blog

.text:00401305                 xor     eax, eax
.text:00401307                 xor     ebx, ebx
.text:00401309                 xor     ecx, ecx
.text:0040130b                 xor     edx, edx
.text:0040130d                 lea     eax, ds:4031b4h
.text:00401313
.text:00401313 loop_comparekey:                                  ; code xref: sub_401109+234j
.text:00401313                 mov     bl, [ecx+eax]
.text:00401316                 mov     dl, savekey1[ecx]
.text:0040131c                 cmp     bl, 0
.text:0040131f                 jz      correct_key
.text:00401325                 add     dl, 5                     ; 再次取计算后的key,如果值+5大于5a的话,值-0d
.text:00401328                 cmp     dl, 5ah
.text:0040132b                 jg      short sub_dl_0d
.text:0040132d
.text:0040132d loc_40132d:                                       ; code xref: sub_401109+23bj
.text:0040132d                 xor     dl, 0ch                   ; if key<&h41 then key=&h4b +loopval
.text:00401330                 cmp     dl, 41h
.text:00401333                 jl      short addmov_dl_4b
.text:00401335                 cmp     dl, 5ah                   ; if key>&h5a then key=&h4b - loopval
.text:00401338                 jg      short submov_dl_4b
.text:0040133a
.text:0040133a loc_40133a:                                       ; code xref: sub_401109+241j
.text:0040133a                                                   ; sub_401109+247j
.text:0040133a                 inc     ecx
.text:0040133b                 cmp     dl, bl                    ; 如果结果不相等则over

本文关键:diablo2002’s crackme1算法分析
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