天书夜读【汇编反C语言】

来源：互联网发布：程序员加班猝死编辑：程序博客网时间：2024/06/10 07:26

感觉作者分析有误！以下是我的分析：

void XXX(a,int b[2][2],int c[2][2],int d[3])

{

ebx = a ;// dword ptr [esp+4]

edx = b ;// dword ptr [esp+8]

ecx = c ; // dword ptr [esp+10h]

esi = d ;// dword ptr [esp+14h]

{

00401000 push ecx

00401001 mov ecx,dword ptr [esp+10h]

00401005 mov edx,dword ptr [esp+8]

00401009 push ebx

0040100A mov ebx,dword ptr [esp+4]

0040100E push esi

0040100 mov esi,dword ptr [esp+14h]

00401013 push ebp

}

int i = 0 ; //eax

{

00401014 xor eax,eax

00401016 push edi

00401017 jmp myunction+20h (401020h)

00401019 lea esp,[esp]

}

int k = 0 ;

{

c[k][0] = d[2]*b[i][1] + d[0]*b[i][0] ;

{

00401020 mov edi,dword ptr [esi+8]

00401023 imul edi,dword ptr [edx+eax*8+4]

00401028 mov ebp,dword ptr [esi]

0040102A imul ebp,dword ptr [edx+eax*8]

0040102E add edi,ebp

00401030 mov dword ptr [ecx],edi

}

ebx = c[k][0] + d[0]*b[i][1] + d[1]*b[i][0] ;

{

00401032 mov edi,dword ptr [esi+0h]

00401035 imul edi,dword ptr [edx+eax*8+4]

0040103A mov ebp,dword ptr [esi+4]

0040103 imul ebp,dword ptr [edx+eax*8]

00401041 add edi,ebp

00401043 mov ebp,dword ptr [ecx]

00401045 add ebp,edi

00401047 add ebx,ebp

}

c[k][1] = d[0]*b[i][1] + d[1]*b[i][0] ;

{

00401049 mov dword ptr [ecx+4],edi

}

i++ ;

k++ ;

{

0040104 inc eax

0040104 add ecx,8

}

00401050 cmp eax,2

00401053 jl myunction+20h (401020h)

}while(i<2)

ebx = ebx +rand();

{

00401055 call rand (401138h)

0040105A add ebx,eax

}

switch(ebx)

{

case: 64h

{

printf("cnt is 100") ;

}

case: 6Eh

{

printf("cnt is 110") ;

break ;

}

deault:

{

printf("“nothing”") ;

}

return ebx ;

}