using void pointer in swap function

Question

I have a swap function like so:

void swap(int i, int j, void* arr[])
{
    void *temp;
    temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
}

I call swap in main like so:

main()
{
    int arr[8] = {4,7,9,2,6,7,8,1};
    void *ptr = arr;
    swap(0, 1, ptr);
    int k;
    for (k=0; k<8; k++)
        printf("%d ", arr[k]);
}

Now, the swap seems to work fine, however instead of swapping 1 value with another, it is swapping 2 values with another 2 values. For example, when I do swap(0, 1, ptr), i get the array

9,2,4,7,6,7,8,1

when I should be getting:

7,4,9,2,6,7,8,1

Instead of swapping 4 and 7, it is swapping 4,7 with 9,2. Why is it doing this?

Answer 1

swap() is treating the array as an array of pointers, but the actual array being passed is an array of type int . Apparently, your system is such that a pointer is the size of two int s, and so everytime it swaps a "pointer", it's really swapping two integers.

You would need your swap routine to be something like this:

void swap(int i, int j, int arr[])
{
    int temp;
    temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
}