Swapping 2 void pointers addresses

Question

I need to create a swap function that takes 2 addresses as input and swaps them regardless what type they point to. Here's my swap function:

void swap(void* x,void* y){
  void* temp=x;
  x=y;
  y=temp;
}

When I use it with integers it works fine and swaps them properly, but using strings the addresses seems to swap inside the function but when I try to call them from outside the function I notice they didn't change at all.

Here's my full code and the resulting output.

  printf("before %s %s\n",(char*)array[i],(char*)array[j] );
  swap(array[i], array[j]);
  printf("after %s %s\n",(char*)array[i],(char*)array[j] );

I casted everything to string to understand what was wrong with them

void swap(void* x,void* y){
  printf("  after IN %s %s\n",(char*)x,(char*)y );
  void* temp=x;
  x=y;
  y=temp;
  printf("  after IN %s %s\n",(char*)x,(char*)y );
}

OUTPUT

before fannullone falafel
  after IN fannullone falafel
  after IN falafel fannullone
after fannullone falafel

Answer 1

To swap two objects in a function you need to pass them to the function by reference.

In C passing by reference means passing objects indirectly through pointers to them. So dereferencing the pointers the function gets a direct access to the original objects and can change them.

So for objects of the type void * the function parameters will have the type void **. The function will look like

void swap( void **x, void **y )
{
    void *temp = *x;
    *x = *y;
    *y = temp;
}

Here is a demonstration program.

#include <stdio.h>

void swap( void **x, void **y )
{
    void *temp = *x;
    *x = *y;
    *y = temp;
}

int main( void ) 
{
    void *s1 = "Hello";
    void *s2 = "World";

    printf( "s1 = %s, s2 = %s\n", ( char * )s1, ( char * )s2 );

    swap( &s1, &s2 );

    printf( "s1 = %s, s2 = %s\n", ( char * )s1, ( char * )s2 );
}

The program output is

s1 = Hello, s2 = World
s1 = World, s2 = Hello