[EXERCISE 0006] Very simple one

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.text:00401000 _main proc near
.text:00401000
.text:00401000 var_8= dword ptr -8
.text:00401000 var_4= dword ptr -4
.text:00401000 argc= dword ptr  8
.text:00401000 argv= dword ptr  0Ch
.text:00401000 envp= dword ptr  10h
.text:00401000
.text:00401000 push    ebp
.text:00401001 mov     ebp, esp
.text:00401003 sub     esp, 8
.text:00401006 mov     [ebp+var_4], 5
.text:0040100D mov     [ebp+var_8], 6
.text:00401014 mov     eax, [ebp+var_8]
.text:00401017 mov     [ebp+var_4], eax
.text:0040101A xor     eax, eax
.text:0040101C mov     esp, ebp
.text:0040101E pop     ebp
.text:0040101F retn
.text:0040101F _main

3 Comments

1. TEHb

   Posted November 19, 2009 at 3:24 PM

   int main()
   { int var_4;
   int var_8;
   var_4 = 5;
   var_8 = 6;
   var_8 = var_4;
   return 0;
   }
2. algemy

   Posted November 20, 2009 at 5:49 PM

   TEHb, I think this program is assigning var_8 to Var_4. Because it is doing a mov of var_8 to eax, and then moving eax into var_4.

   int main( int *argc, int *argv)
   {
   int var_8, var_4;
   var_4 = 5;
   var_8 = 6;
   var_4 = var_8 ;
   return 0;
   }
3. algemy

   Posted November 20, 2009 at 5:50 PM

   BTW, argv should have been char *argv not int, my mistake.