Using pointers to swap int array values

Question

I am supposed to use pointers to swap ints in an array. It compiles with no errors or warnings and runs but does not swap the ints. Any suggestions would be helpful!!!

Here is the tester:

#import <stdio.h>

void swap( int ary[] );

int main(  int argc, char*argv[] )
{
    int ary[] = { 25, 50 };
    printf( "The array values are: %i and %i \n", ary[0], ary[1] );
    swap( ary );
    printf( "After swaping the values are: %i and %i \n", ary[0], ary[1] );

    return 0;
}

Here is the swap function:

void swap( int ary[] )
{
    int temp = *ary;
    *ary = *(ary + 1);
    *ary = temp;
}

This is what is displayed after running:

The array values are: 25 and 50
After swaping the values are: 25 and 50

Answer 1

I hate spoiling this but it looks like a typo more than anything.

In your swap function:

*ary = temp;

should be:

*(ary + 1) = temp;

edit: Is there a reason you're not using array notation? I think it's a bit clearer for things like this:

int temp = ary[0];
ary[0] = ary[1];
ary[1] = temp;

Answer 2

Examine your swap function more carefully:

void swap( int ary[] )
{
    int temp = *ary;
    *ary = *(ary + 1);
    *ary = temp;
}

When does *(ary + 1) get assigned to?

Answer 3

You move the second value into the first spot, and then move the first value back into the first spot.