Question

My program is written in C++ with some x86 assembly language being embedded into it. I have 2 nested for-loops in assembly language that I must go through. However, however when I compile my program, I get an infinite loop. The equivalent program in C++ would look something like this:

#include<iostream>
using namespace std;
int main()
{
    int a[4] = {3,6,4,7};

    for(int k = 0 ; k < 4;k++)
    {
        for(int l = 0 ; l < a[k];l++)
        {
            cout<<'*';
        }
        cout<<endl;
    }

    system("pause");
    return 0;
}

 /*

  *** 
  ******
  ****
  *******
  Press any key to continue . . .
  */

This is the same thing, but done with assembly mixed in.

#include<iostream>
using namespace std;
void output(); //function for making an '*'
void makeSpace(); //function for making a space

int main()
{
    int a[4]={3,6,4,7};

    int counter = 0; //counter that will be used for first forloop
    int counter2 = 0; // counter to be used for second forloop

_asm{

    mov ebx,0 // this is to move from element 0,1,2,3,4, through the array
    mov ecx,0 // ecx will get the data from the array, only to be used as a
              // counter in forloop2 though.

     for1:

    cmp counter,4 //begins for firloop

    je starts 
    mov ecx,[a+ebx] // move the 0th element from array to ecx
    add ebx,4// ebx = ebx+4, I'm doing this to advance the array position (int)
    inc counter// increment counter by one

      for2:
    cmp counter2,ecx //begin forloop2, 
    je starts2
    call output
    inc counter2 //increment counter2 by one

    jmp for2 
      starts2:

    call makeSpace
    jmp for1
      starts:
}

    return 0;
}

void output()
{
    cout<<'*';
    }

void makeSpace()
{
    cout<<endl;
}

Why does this result in an infinite loop?

Answer 1

There are at least two things you need to fix:

• When you call output() only the following registers are guaranteed to not be trashed:

  • edi, esi, ebx, and ebp

  In particular, you’re using ecx, which the function is permitted to trash.

• You never reset counter2 to 0, so the inner loop isn’t equivalent to your C code.