[EXERCISE 0008] WTF? Where is my modulo?

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int main(int argc, char* argv[])
{
	int a;
	a = 11 % 3;
	return 0;
}

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.text:00401000 _main proc near
.text:00401000
.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 push    ecx
.text:00401004 mov     [ebp+var_4], 2
.text:0040100B xor     eax, eax
.text:0040100D mov     esp, ebp
.text:0040100F pop     ebp
.text:00401010 retn
.text:00401010 _main endp

5 Comments

1. TEHb

   Posted November 20, 2009 at 3:03 PM

   preprocessor works :)
2. Baramine

   Posted November 20, 2009 at 3:16 PM

   GCC does it also on linux :)
3. algemy

   Posted November 20, 2009 at 4:24 PM

   I am still a newby on this, but it seems to me that the assembly code does not shows how the calculation (11%3) is being done. Instead it is moving the result (11%3=2) to variable a (ebp+var_4).

   Thus, I am assuming that when the code is compile, VS automatically generates the result for this simple calculation.
4. Steve

   Posted November 21, 2009 at 4:44 PM

   Yes, algemy, it’s because the compiler optimized the simple math calculation to reduce the amount of instructions and therefore making the program more efficient.
5. Slow

   Posted December 12, 2009 at 5:11 AM

   Yeah what Steve said, optimization took out some instructions. Further instructions remove the variable size build up and tear down and even further optimization turns this whole function into 0×90 :-)