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[求助]FGInt RSA 问题
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发表于: 2008-9-21 00:41
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FGInt RSA 问题
有一程序PEID插件CryptoANALyzer 检测到Base10StringToFGInt和RSADecrypt
因为:
查相关资料得RSADecrypt原型:
RSADecrypt(test, d, n, Nilgint, Nilgint, Nilgint, Nilgint, test);
其中test是C和M
共8个参数
又
fastcall:
是Delphi默认所使用的参数传递方式,此种方式在传递参数时把前三个参数放在CPU的EAX,EDX,ECX三个缓存器种,剩下的参数则会由左到右地被取出放到堆栈中,而当被调用的函数结束之 后,则是由被调用函数来清除堆栈上的参数数据.
所以....
代码:
0051BF63 . 8D45 D8 lea eax, dword ptr [ebp-28] 4 nul
0051BF66 . 50 push eax
0051BF67 . 8D45 D8 lea eax, dword ptr [ebp-28] 5 nul
0051BF6A . 50 push eax
0051BF6B . 8D45 D8 lea eax, dword ptr [ebp-28] 6 nul
0051BF6E . 50 push eax
0051BF6F . 8D45 D8 lea eax, dword ptr [ebp-28] 7 nul
0051BF72 . 50 push eax
0051BF73 . 8D45 F4 lea eax, dword ptr [ebp-C] 8 test 结果,即M
0051BF76 . 50 push eax
0051BF77 . 8D4D E0 lea ecx, dword ptr [ebp-20]
0051BF7A . 8D55 E8 lea edx, dword ptr [ebp-18]
0051BF7D . 8B45 F0 mov eax, dword ptr [ebp-10]
0051BF80 . E8 EFFBFEFF call <RsaDecrypt>
寄存器:
EAX 00EB2EA8 1 test 即C
ECX 0012FDD8 3 D
EDX 0012FDE0 2 N
EBX 0043D76C b.0043D76C
ESP 0012FBC0
EBP 0012FDF8
ESI 00EA17C8
EDI 00516AC0 b.00516AC0
EIP 0051BF80 b.0051BF80
不知以上分析是否正确，忘指教 :)
看其值:
C/eax=[00eb2ea8]
00EB2EA8 3F A8 4C 5C 53 D3 F8 F2 86 DA 59 98 78 29 CA BE ?↙\S峪騿赮榵)示
00EB2EB8 2D 30 8D 3A C1 DF 82 15 D9 DA 20 BA 5B 2C 2C 43 -0?吝?仝 篬,,C
00EB2EC8 00 76 52 00 0C 76 52 00 B8 0A EB 00 30 11 00 00 .vR..vR.??0..
C=3FA84C5C53D3F8F286DA59987829CABE2D308D3AC1DF8215D9DA20BA5B2C2C43
D/ecx=[0012FDD8]
0012FDD8 01 00 00 00 0C 1B EB 00 01 00 00 00 68 32 EA 00 ....?...h2?
0012FDE8 A8 2E EB 00 D8 CD 51 00 64 2B EA 00 64 2B EA 00 ??赝Q.d+?d+?
C=1 //?????这里是不是有问题？
N/edx=[0012FDE0]
0012FDE0 01 00 00 00 68 32 EA 00 A8 2E EB 00 D8 CD 51 00 ...h2???赝Q.
0012FDF0 64 2B EA 00 64 2B EA 00 1C FE 12 00 87 18 47 00 d+?d+? ?.?G.
N=1 //...
RSADecrypt后得到明文M
00EB2E88 55 6E 52 65 67 69 73 74 65 72 20 56 65 72 73 69 UnRegister Versi
00EB2E98 6F 6E 00 00 32 00 00 00 01 00 00 00 20 00 00 00 on..2...... ...
M="UnRegister Version"
前面还有一段代码，觉得有点可疑
Base10StringToFGInt()
333335333939333634343735373034383137303933393439373735383035343732353832363033
Base10StringToFGInt()
3633343632333337323731343932303238373136383536313837333630333930393637363130393931333833393837373530363235393934383134313230383431313637343836393332313337
难道这两个才是D和M?但长度相差...
还有一个问题就是如果确定这个是RSA几？RSA128?RSA93?等等...
对RSA不了解呵
指点迷津 不胜感激
有一程序PEID插件CryptoANALyzer 检测到Base10StringToFGInt和RSADecrypt
因为:
查相关资料得RSADecrypt原型:
RSADecrypt(test, d, n, Nilgint, Nilgint, Nilgint, Nilgint, test);
其中test是C和M
共8个参数
又
fastcall:
是Delphi默认所使用的参数传递方式,此种方式在传递参数时把前三个参数放在CPU的EAX,EDX,ECX三个缓存器种,剩下的参数则会由左到右地被取出放到堆栈中,而当被调用的函数结束之 后,则是由被调用函数来清除堆栈上的参数数据.
所以....
代码:
0051BF63 . 8D45 D8 lea eax, dword ptr [ebp-28] 4 nul
0051BF66 . 50 push eax
0051BF67 . 8D45 D8 lea eax, dword ptr [ebp-28] 5 nul
0051BF6A . 50 push eax
0051BF6B . 8D45 D8 lea eax, dword ptr [ebp-28] 6 nul
0051BF6E . 50 push eax
0051BF6F . 8D45 D8 lea eax, dword ptr [ebp-28] 7 nul
0051BF72 . 50 push eax
0051BF73 . 8D45 F4 lea eax, dword ptr [ebp-C] 8 test 结果,即M
0051BF76 . 50 push eax
0051BF77 . 8D4D E0 lea ecx, dword ptr [ebp-20]
0051BF7A . 8D55 E8 lea edx, dword ptr [ebp-18]
0051BF7D . 8B45 F0 mov eax, dword ptr [ebp-10]
0051BF80 . E8 EFFBFEFF call <RsaDecrypt>
寄存器:
EAX 00EB2EA8 1 test 即C
ECX 0012FDD8 3 D
EDX 0012FDE0 2 N
EBX 0043D76C b.0043D76C
ESP 0012FBC0
EBP 0012FDF8
ESI 00EA17C8
EDI 00516AC0 b.00516AC0
EIP 0051BF80 b.0051BF80
不知以上分析是否正确，忘指教 :)
看其值:
C/eax=[00eb2ea8]
00EB2EA8 3F A8 4C 5C 53 D3 F8 F2 86 DA 59 98 78 29 CA BE ?↙\S峪騿赮榵)示
00EB2EB8 2D 30 8D 3A C1 DF 82 15 D9 DA 20 BA 5B 2C 2C 43 -0?吝?仝 篬,,C
00EB2EC8 00 76 52 00 0C 76 52 00 B8 0A EB 00 30 11 00 00 .vR..vR.??0..
C=3FA84C5C53D3F8F286DA59987829CABE2D308D3AC1DF8215D9DA20BA5B2C2C43
D/ecx=[0012FDD8]
0012FDD8 01 00 00 00 0C 1B EB 00 01 00 00 00 68 32 EA 00 ....?...h2?
0012FDE8 A8 2E EB 00 D8 CD 51 00 64 2B EA 00 64 2B EA 00 ??赝Q.d+?d+?
C=1 //?????这里是不是有问题？
N/edx=[0012FDE0]
0012FDE0 01 00 00 00 68 32 EA 00 A8 2E EB 00 D8 CD 51 00 ...h2???赝Q.
0012FDF0 64 2B EA 00 64 2B EA 00 1C FE 12 00 87 18 47 00 d+?d+? ?.?G.
N=1 //...
RSADecrypt后得到明文M
00EB2E88 55 6E 52 65 67 69 73 74 65 72 20 56 65 72 73 69 UnRegister Versi
00EB2E98 6F 6E 00 00 32 00 00 00 01 00 00 00 20 00 00 00 on..2...... ...
M="UnRegister Version"
前面还有一段代码，觉得有点可疑
Base10StringToFGInt()
333335333939333634343735373034383137303933393439373735383035343732353832363033
Base10StringToFGInt()
3633343632333337323731343932303238373136383536313837333630333930393637363130393931333833393837373530363235393934383134313230383431313637343836393332313337
难道这两个才是D和M?但长度相差...
还有一个问题就是如果确定这个是RSA几？RSA128?RSA93?等等...
对RSA不了解呵
指点迷津 不胜感激
[注意]传递专业知识、拓宽行业人脉——看雪讲师团队等你加入！
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啥?你是说FGInt 还是说升旗?
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DiKeN兄的意思是否:
寄存器: EAX 00EB2EA8 1 test 即C ECX 0012FDD8 3 D EDX 0012FDE0 2 N EBX 0043D76C b.0043D76C ESP 0012FBC0 EBP 0012FDF8 ESI 00EA17C8 EDI 00516AC0 b.00516AC0 EIP 0051BF80 b.0051BF80 即  但指向的数据是  其实我的目的就是能从这个RSADecrypt(M=C^D mod N)找出D和N来 :) 理解能力有限，多多指教啊！ |
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对了[Number Pointer] + 1= [Number Pointer - 4] ->5+1=6
 可是为什么呢？你有FGInt的想关资料吗？比连源码或库 我上网找没找着....呵 zq的大小你是如何确定的？根据FGInt? 那我这里 zq[]="/x05/x00/x00/x00/x00/x00/x00/x00/xCB/x2B/x68/x21/x00/x00/x00/x00/xCB/x1B/x34/x09"; 是吗？ 那ECX即D呢？那里会得到什么？ 09 00 00 00 00 00 00 00 A9 54 85 78 00 00 00 00 1F 9D F4 2F 11点了要睡觉啦写不赢啦 我还是不知道如何得到D，N DikeN继续指点:) |
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几天没上网，没想DikeN还是那么有耐心
上次发完贴后就逛了一下，最后在submanifold.be找到了FGInt，和DiKeN说的一样 :) 然后又下载了个DELPHI 7精简版。 以前没用过delphi,这次也算是上手了呵 终于有点头绪了 : ) 主要就是弄明白FGInt的结构:) FGInt { Sign:TSign; Number:Array of Cardinal; //可以由Length(fgint.Number)来取得元素个数 } 其实挺简单的... 0051BE5C . 8D55 E8 lea edx, dword ptr [ebp-18] ;335399364475704817093949775805472582603 0051BE5F . A1 80815200 mov eax, dword ptr [528180] 0051BE64 . E8 7FD8FEFF call <Base10StringToFGInt> 0051BE69 . 8D55 E0 lea edx, dword ptr [ebp-20] ;63462337271492028716856187360390967610991383987750625994814120841167486932137 0051BE6C . A1 84815200 mov eax, dword ptr [528184] 0051BE71 . E8 72D8FEFF call <Base10StringToFGInt> Delphi代码如下 program Project1; {$APPTYPE CONSOLE} uses Windows, Messages, FGInt, FGIntRSA; Var n,e: TFGInt; name,sn: String; size,i:Integer; Begin Base10StringToFGInt('335399364475704817093949775805472582603', n); size:=SizeOf(n); writeln('Size of n:',size); size:=SizeOf(TFGInt); writeln('Size of TFGInt:',size); size:=SizeOf(n.Sign); writeln('Size of Sign:',size); size:=SizeOf(n.Number); //写这些时还没看帮助，这时还不知道Length可以取得array of的长度 :) writeln('Size of Number:',size); size:=Length(n.Number); //length可以取得 writeln('Lenght of n.Number:',size); for i:=0 to size do Writeln(i,':',n.Number[i]); Base10StringToFGInt('63462337271492028716856187360390967610991383987750625994814120841167486932137', n); size:=SizeOf(n); writeln('Size of n:',size); size:=SizeOf(TFGInt); writeln('Size of TFGInt:',size); size:=SizeOf(n.Sign); writeln('Size of Sign:',size); size:=SizeOf(n.Number); writeln('Size of Number:',size); size:=Length(n.Number); writeln('Lenght of n.Number:',size); for i:=0 to size do Writeln(i,':',n.Number[i]); End; OD里调试我的程序，看到内存里  目标程序里的则是  为什么会这样？ 问题W：这里n.Number的每个元素都会空一个DWORD?? 有 D=335399364475704817093949775805472582603 N=63462337271492028716856187360390967610991383987750625994814120841167486932137 M=3FA84C5C53D3F8F286DA59987829CABE2D308D3AC1DF8215D9DB30BA5B2C2C43 So C=???? 最后C得出乱码... 这种情况是因为目标程序的Base10StringToFGInt有变形是吗？ 几天没想这个问题了，下午放学回来再弄弄，上课去了 再次感谢DikeN :) |
