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Home/ Questions/Q 8478007
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Editorial Team
  • 0
Editorial Team
Asked: June 10, 20262026-06-10T18:37:31+00:00 2026-06-10T18:37:31+00:00

In the below code, output remains same in both cases,thanks for pointing what am

  • 0

In the below code, output remains same in both cases,thanks for pointing what am I missing:-

Before Swap:-
a=10     b=512
After Swap:-
a=10     b=512

Following is the code, It compiles and runs without any problem:-

#include <stdio.h>
int swap(int* x, int* y)
{
    if(x != y)
    {
        _asm
        {
            mov eax,[x]; x into eax
            mov ebx,[y]
            mov [x],ebx;swapping now
            mov [y],eax
        }
    }
    return 0;
}

int main () {
  int a=10,b=512;
  printf("Before Swap:- \na=%d\t b=%d\n",a,b);
  swap(&a,&b); 
  printf("After Swap:- \na=%d\t b=%d",a,b);//Value remains same
  return 0;
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-10T18:37:33+00:00Added an answer on June 10, 2026 at 6:37 pm

    No indirection on variables inside assembly block wont work.Instead take addresses in registers and then only try indirection.It will rather break into something like mov eax, DWORD PTR _x$[ebp]

    #include <stdio.h>
    int swap(int* x, int* y)
    {
        if(x != y)
        {
            _asm
            {
                mov eax,x
                mov ebx,y
                mov ecx,[ebx]
                xchg ecx,[eax]
                xchg [ebx],ecx
            }
        }
        return 0;
    }
    
    int main () {
      int a=10,b=512;
      printf("Before Swap:- \na=%d\t b=%d\n",a,b);
      swap(&a,&b);
      printf("After Swap:- \na=%d\t b=%d",a,b);
      getchar();
      return 0;
    }
    
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