Thread: Really really easy, dumb question

Generally, you don't.

Pass in the array you want to fill in, and everything will work better.

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Mats

It's difficult and messy to do so. Instead you need to use pointers. Now hold on, it's not as bad as it seems. Here, lemme show you:

Code:

void reverse_list(int *temp)                /*creates a list of ints from 0 to N*/
{ /* Incidentally (int temp[]) should work just the same.

    int N = 20;
    int temp[N];
    
    for(int i = 0; i < N; i++)
    {
            temp[N - i] = i;
    }
   /* no need to return because you're manipulating the array directly through pointers.
}

Then to use it:

Code:

int array[20];

/* fill the array and then */
reverse_list (array);

Now, that fixes the code, but there's still a big problem. It won't work. i mean it'll pass the array and whatnot, but it won't reverse the array. Instead it will fill the array with the numbers 20 to 1 regardless of what was in there. However, I leave that to you to fix.

Originally Posted by Hexadakota

Does that mean that occasionally you can?

Yes, there are some special cases where you can return a pointer or reference to an array (or an actual array, but that's extremely rare indeed). Just make sure that the pointer or reference you return points to data that is exists beyond the return from the function:
1. static local variable.
2. global variable.
3. passed in parameter.
4. memory allocated with new/malloc.

In case 4, you must also remember to free the memory afterwards (using delete or free respectively).

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Mats