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注册时间2007-12-2
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周游历练
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发表于 2008-3-8 19:31:16
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哈哈,爆破了 可以用的没有什么限制了,拿出我的爆破破解过程。
esp定律脱壳后,是Delphi程序,试运行程序看到标题有未注册字样。我们就拿它开刀。
OD载入找字符串,找到未注册字样。来到下面的代码:
00594947 64:FF30 push dword ptr fs:[eax]
0059494A 64:8920 mov dword ptr fs:[eax],esp
0059494D 8B45 FC mov eax,dword ptr ss:[ebp-4]
00594950 E8 27CEF7FF call dumped_.0051177C /关键call,F7跟进,又出现了个call继续跟进
00594955 84C0 test al,al
00594957 74 1C je short dumped_.00594975
00594959 8D45 F4 lea eax,dword ptr ss:[ebp-C]
0059495C B9 F0495900 mov ecx,dumped_.005949F0 ; [已注册]
00594961 8B55 F8 mov edx,dword ptr ss:[ebp-8]
00594964 E8 BB0AE7FF call dumped_.00405424
00594969 8B55 F4 mov edx,dword ptr ss:[ebp-C]
0059496C 8BC3 mov eax,ebx
0059496E E8 856FF2FF call dumped_.004BB8F8
00594973 EB 1F jmp short dumped_.00594994
00594975 8D45 F0 lea eax,dword ptr ss:[ebp-10]
00594978 B9 044A5900 mov ecx,dumped_.00594A04 ; [未注册]
0059497D 8B55 F8 mov edx,dword ptr ss:[ebp-8]
00594980 E8 9F0AE7FF call dumped_.00405424
00594985 8B55 F0 mov edx,dword ptr ss:[ebp-10]
00594988 8BC3 mov eax,ebx
跟进call后,来到下面的代码:
005119DF 8BC6 mov eax,esi
005119E1 E8 620B0000 call dumped_.00512548
005119E6 EB 02 jmp short dumped_.005119EA
005119E8 33DB xor ebx,ebx //把xor ebx,ebx 改成or ebx ebx
005119EA 33C0 xor eax,eax
005119EC 5A pop edx ; 0012FDD0
005119ED 59 pop ecx
005119EE 59 pop ecx
005119EF 64:8910 mov dword ptr fs:[eax],edx
005119F2 68 0C1A5100 push dumped_.00511A0C
005119F7 8D45 F4 lea eax,dword ptr ss:[ebp-C]
005119FA BA 03000000 mov edx,3
005119FF E8 1C37EFFF call dumped_.00405120
00511A04 C3 retn
00511A05 ^ E9 662FEFFF jmp dumped_.00404970
00511A0A ^ EB EB jmp short dumped_.005119F7
00511A0C 8BC3 mov eax,ebx
00511A0E 5E pop esi
00511A0F 5B pop ebx
00511A10 8BE5 mov esp,ebp
通把xor ebx ebx改成or ebx ebx ,达到了爆破。
不知道你明白没有 ,哈哈 |
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