PC 控制台(PC Controler) V1.5.75 破解过程 (5千字) -pc6资讯

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PC 控制台(PC Controler) V1.5.75 破解过程 (5千字) 时间：2004/10/15 0:54:00来源：本站整理作者：蓝点我要评论(0): 　

PC 控制台(PC Controler) V1.5.75 破解过程

破解撰写：leeyam[BCG]

首先运行该程序，随意填写注册信息，提示需要重新启动软件验证注册码！
判断该程序先将输入的注册码存放某个位置，然后启动时调用！

用Language发现是用ASPACK加的壳，很容易脱掉。然后利用W32Dasm反编，查找字串，发现程序会调用注册表，"\Software\MySoftware Studio\UserInfo"
从第一个调用入手：
* Possible StringData Ref from Code Obj ->"\Software\MySoftware Studio\UserInfo"
|
:00489B6C BA909E4800 mov edx, 00489E90
:00489B71 8B45F8 mov eax, dword ptr [ebp-08]
:00489B74 E81B78FDFF call 00461394
:00489B79 8845FF mov byte ptr [ebp-01], al
:00489B7C 807DFF00 cmp byte ptr [ebp-01], 00
:00489B80 750A jne 00489B8C
:00489B82 E841A7F7FF call 004042C8
:00489B87 E9C3020000 jmp 00489E4F

* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:00489B80(C)
|

* Possible StringData Ref from Code Obj ->"UserName"
|
:00489B8C BAC09E4800 mov edx, 00489EC0
:00489B91 8B45F8 mov eax, dword ptr [ebp-08]
:00489B94 E8A77FFDFF call 00461B40
:00489B99 84C0 test al, al
:00489B9B 0F848D020000 je 00489E2E

* Possible StringData Ref from Code Obj ->"UserID"
|
:00489BA1 BAD49E4800 mov edx, 00489ED4
:00489BA6 8B45F8 mov eax, dword ptr [ebp-08]
:00489BA9 E8927FFDFF call 00461B40
:00489BAE 84C0 test al, al
:00489BB0 0F8478020000 je 00489E2E

* Possible StringData Ref from Code Obj ->"UserEmail"
|
:00489BB6 BAE49E4800 mov edx, 00489EE4
:00489BBB 8B45F8 mov eax, dword ptr [ebp-08]
:00489BBE E87D7FFDFF call 00461B40
:00489BC3 84C0 test al, al
:00489BC5 0F8463020000 je 00489E2E

* Possible StringData Ref from Code Obj ->"10674538806"
|
:00489BCB BAF89E4800 mov edx, 00489EF8
:00489BD0 8B45F8 mov eax, dword ptr [ebp-08]
:00489BD3 E8687FFDFF call 00461B40
:00489BD8 84C0 test al, al
:00489BDA 0F844E020000 je 00489E2E
:00489BE0 33C0 xor eax, eax
:00489BE2 55 push ebp
:00489BE3 68389C4800 push 00489C38
:00489BE8 64FF30 push dword ptr fs:[eax]
:00489BEB 648920 mov dword ptr fs:[eax], esp
:00489BEE 8D4DF4 lea ecx, dword ptr [ebp-0C]

* Possible StringData Ref from Code Obj ->"UserName"
|
:00489BF1 BAC09E4800 mov edx, 00489EC0
:00489BF6 8B45F8 mov eax, dword ptr [ebp-08]
:00489BF9 E8167DFDFF call 00461914
:00489BFE 8D4DF0 lea ecx, dword ptr [ebp-10]

* Possible StringData Ref from Code Obj ->"UserID"
|
:00489C01 BAD49E4800 mov edx, 00489ED4
:00489C06 8B45F8 mov eax, dword ptr [ebp-08]
:00489C09 E8067DFDFF call 00461914
:00489C0E 8D4DEC lea ecx, dword ptr [ebp-14]

* Possible StringData Ref from Code Obj ->"UserEmail"
|
:00489C11 BAE49E4800 mov edx, 00489EE4
:00489C16 8B45F8 mov eax, dword ptr [ebp-08]
:00489C19 E8F67CFDFF call 00461914

* Possible StringData Ref from Code Obj ->"10674538806"
|
:00489C1E BAF89E4800 mov edx, 00489EF8
:00489C23 8D4DE8 lea ecx, dword ptr [ebp-18]
:00489C26 8B45F8 mov eax, dword ptr [ebp-08]
:00489C29 E8E67CFDFF call 00461914
:00489C2E 33C0 xor eax, eax
:00489C30 5A pop edx
:00489C31 59 pop ecx
:00489C32 59 pop ecx
:00489C33 648910 mov dword ptr fs:[eax], edx
:00489C36 EB1D jmp 00489C55
:00489C38 E9F3A2F7FF jmp 00403F30
:00489C3D C645FF00 mov [ebp-01], 00
:00489C41 E852A6F7FF call 00404298
:00489C46 E87DA6F7FF call 004042C8
:00489C4B E9FF010000 jmp 00489E4F
:00489C50 E843A6F7FF call 00404298

以上可以很容易发现是在调入注册表里填写的注册信息！相信判断注册码的位置就在下面，再往下看看！发现以下代码想必是判断关键：
* Referenced by a (U)nconditional or (C)onditional Jump at Address:
|:00489D06(C)
|
:00489D49 8B45E8 mov eax, dword ptr [ebp-18]…………………………调入假码
:00489D4C 8B55DC mov edx, dword ptr [ebp-24]…………………………调入真码
:00489D4F E864AEF7FF call 00404BB8…………………………判断注册码！
:00489D54 0F9445FF sete byte ptr [ebp-01]
:00489D58 807DFF00 cmp byte ptr [ebp-01], 00
:00489D5C 0F84D0000000 je 00489E32

于是调用我的TRW2000，直接下中断 bpx 489d4f 然后运行该程序，顺利被拦截，马上 D edx 得到正确注册码，添入软件后注册成功！

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