Swapping Array elements call by reference

Question

I had to swap array elements with call by reference to reduce the lines of code in my bigger project, and I am successful in swapping them but this code gives a bunch of warnings, how do I do this efficiently?

#include <stdio.h>
void swap(int *arr, int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
}
int main(void)
{
    int a[5] = {2, 4, 6, 2, 3};
    swap(&a, 1, 2);
    for (int i = 0; i < 5; i++)
    {
        printf("%d", a[i]);
    }

    return 0;
}

OUTPUT =>

swap_array.c: In function 'main':
swap_array.c:11:10: warning: passing argument 1 of 'swap' from incompatible pointer type [-Wincompatible-pointer-types]
   11 |     swap(&a, 1, 2);
      |          ^~
      |          |
      |          int (*)[5]
swap_array.c:2:16: note: expected 'int *' but argument is of type 'int (*)[5]'
    2 | void swap(int *arr, int i, int j)
      |           ~~~~~^~~
26423

Answer 1

You declared an object of the array type int [5]

int a[5] = {2, 4, 6, 2, 3};

So the expression &a has the pointer type int ( * )[5]. But the function expects the first argument of the type int *

void swap(int *arr, int i, int j)

There is no implicit conversion from the type int ( * )[5] to the type int *. So the compiler issues an error message.

Instead you need to call the function like

swap(a, 1, 2);

In this case the array designator a is implicitly converted to pointer to its first element of the type int * and using the pointer arithmetic you can access elements of the array with the indices 1 and 2.

However it is better to declare and define the function only with two parameters the following way

void swap( int *px, int *py )
{
    int temp = *px
    *px = *py;
    *py = temp;
}

and call the function like

swap( a + 1, a + 2 );

In C passing by reference means passing objects indirectly through pointers to them. Thus dereferencing the pointers you can have a direct access to the objects pointed to by the pointers.

Pay attention to that for example the expression a[i] is evaluated like *( a + i ) where the array designator a is converted to pointer to its first element.

Answer 2

Your have an extra & operator, just remove it.

swap(a, 1, 2);

Arrays, like your a variable, decay to pointers in C, so when you do &a, you're creating a double pointer, which your function doesn't accept.

You can read more about pointer decay here.