Working with pointer : pointer assignment to pointer

Question

Here I have tried to make a function which may swap two values without a third variable. What other implementation might work to do so other than the one stated below?

void swap(int* x, int* y)
{
    (*x)=(*x)+(*y);
    (*y)=(*x)-(*y);
    (*x)=(*x)-(*y);

}

@OliCharlesworth what is wrong with the application? Why is the above code invalid? — Priyanka, Jun 22 '14 at 08:42
It *is* valid, at least, it won't cause those error messages. — Oliver Charlesworth, Jun 22 '14 at 08:44
There are cases where the result of the operation overflows. — BLUEPIXY, Jun 22 '14 at 12:35
possible duplicate of [How does XOR variable swapping work?](http://stackoverflow.com/questions/249423/how-does-xor-variable-swapping-work) — Patrick M, Jul 01 '14 at 17:27

score 3 · Answer 1 · answered Jun 22 '14 at 08:10

3

It looks like you are trying to do the XOR swap trick?

#include <stdio.h>

void swap(int* x, int* y)
{
    *x^=*y;
    *y^=*x;
    *x^=*y;
}
int main(){
    int x = 1;
    int y = 2;
    swap(&x, &y);
    printf("x = %d, y =%d\n", x, y);
}

answered Jun 22 '14 at 08:10

merlin2011

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Nice example! Simple and concise – ALM865 Jun 22 '14 at 08:13
2

Note: Please don't ever do this in code. – chris Jun 22 '14 at 08:13
It's OK to do in code if you use `unsigned int`. – M.M Jun 22 '14 at 11:23
Thanks @merlin2011 for introducing me to something new (atleast for me). – Priyanka Jun 22 '14 at 14:22

score 0 · Answer 2 · answered Jun 22 '14 at 10:38

0

it should work actually...

I used the following code with your function:

void swap(int* x, int* y)
{
  (*x)=(*x)+(*y);
  (*y)=(*x)-(*y);
  (*x)=(*x)-(*y);
}

int a1=1;
int a2=2;

main()
{
   swap(&a1,&a2);

   printf("a1= %d\na2= %d \n",a1 ,a2);
}

answered Jun 22 '14 at 10:38

aalbashar

1

I had tried using the same code earlier but it didn't work well at that time but now it works fines. May be it was some interpreter etc issue. @Oli Charlesworth sorry for the inconvenience. – Priyanka Jun 22 '14 at 14:19

Working with pointer : pointer assignment to pointer

2 Answers2