Question

If this is possible:

#include <stdio.h>
#include <process.h>

#define SIZE 5

void PassingArray(int arr[])
{
    int i=0;
    for(i=0 ; i<SIZE ; i++)
    {
        printf("%d, ", arr[i]);
    }
    printf("\n");
}

main()
{
    int myIntArray[5] = {1, 2, 3, 4, 5};

    PassingArray(myIntArray);

    system("PAUSE");
}

Then why the following is illegal?

#include <stdio.h>
#include <process.h>

#define SIZE 5

int ReturningArray()[]
{
    int myIntArray[5] = {1, 2, 3, 4, 5};

    return myIntArray;
}

main()
{
    int myArray[] = ReturningArray();

    system("PAUSE");
}

Answer 1

There’s multiple reasons why this doesn’t work.

The first is simply that it’s prohibited by the language – the return type of a function shall not be an array (it also can’t be a function).

The second is that even if you were allowed to declare ReturningArray as you do, you could never write a valid return statement in that function – an expression with array type that is not the subject of the unary & or sizeof operators evaluates to a pointer to the first element of the array, which no longer has array type. So you can’t actually make return see an array.

Thirdly, even if we somehow had a function returning an array type, you couldn’t use that return value as the initialiser of an array variable – the return value would again evaluate to a pointer to the first element of the array: in this case a pointer to int, and a pointer to int isn’t a suitable initialiser for an array of int.