Unable to swap values

Question

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;
}

Perhaps relevant: http://stackoverflow.com/questions/11529778/reason-why-custom-loop-is-faster-bad-compiler-unsafe-custom-code-lucklucky — sehe, Sep 02 '12 at 17:01
Checking for x != y probably costs more than just exchanging two identical values. If this is an exercise for learning assembly code, fine; but the compiler will do just as good a job as you can here (in fact, better -- there's an unneeded instruction in the assembly code). — Pete Becker, Sep 02 '12 at 19:03

score 4 · Accepted Answer · answered Sep 02 '12 at 16:59

4

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;
}

answered Sep 02 '12 at 16:59

perilbrain

7,961
2
27
35

It's not wrong, but could you please not advocate `xchg` with a memory operand? – harold Sep 02 '12 at 17:02
4

@harold:- long days gone,being with assembly. So just dropped the ash of burnt up knowledge with mere perfection. :) – perilbrain Sep 02 '12 at 17:05
swap without using third variable in c language : http://allitstuff.com/swap-without-using-third-variable-in-c-language/ – jaydeep namera Aug 21 '13 at 09:08

score -3 · Answer 2 · answered Sep 02 '12 at 17:30

-3

You can swap with Xor operation -

void swap(int *x, int *y)
{
     *x = *x ^ *y; /* #1 */
     *y = *x ^ *y; /* #2 */
     *x = *x ^ *y; /* #3 */
}

answered Sep 02 '12 at 17:30

URL87

10,667
35
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5

That destroys the value if both pointers are equal, and it is likely slower than a standard swap using a temporary variable. – Daniel Fischer Sep 02 '12 at 17:32
5

The question wasn't how to swap, it was why the given assembly code isn't working. And XOR swap is worse than normal swap in almost every way. – interjay Sep 02 '12 at 17:32
1

2 downvote ... at least now I know that this is not so good patent. – URL87 Sep 02 '12 at 17:50
1

I gave it a 3rd down vote. Now if you delete it you get a badge. `` – Pete Becker Sep 02 '12 at 19:00

Unable to swap values

2 Answers2