CODEWITHSUNDEEP
carraysreturn
user366312
user366312

Reputation: 16928

Why is this array notation illegal in C?

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");
}

Upvotes: 1

Views: 863

Answers (5)

cpx
cpx

Reputation: 17567

You can't return array from a function, but It is possible that you can declare a function returning a (reference in C++) or pointer to array as follows:

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

int (*ReturningArray())[sizeof(myIntArray)/sizeof(int)] {

    return &myIntArray;
}

Upvotes: 0

caf
caf

Reputation: 239061

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.

Upvotes: 2

Steve Jessop
Steve Jessop

Reputation: 279265

PassingArray is legal, but it does not pass an array. It passes a pointer to the first element of an array. void PassingArray(int arr[]) is a confusing synonym for void PassingArray(int *arr). You can't pass an array by value in C.

ReturningArray is not allowed, you can't return an array by value in C either. The usual workaround is to return a struct containing an array:

typedef struct ReturnArray {
    int contents[5];
} ReturnArray;

ReturnArray ReturningArray()
{
    ReturnArray x = {{1, 2, 3, 4, 5}};
    return x;
}

Arrays are second-class citizens in C, the fact that they can't be passed or returned by value is historically related to the fact that they can't be copied by assignment. And as far as I know, the reason for that is buried in the early development of C, long before it was standardized, when it wasn't quite decided how arrays were going to work.

Upvotes: 2

halfdan
halfdan

Reputation: 34214

There are several problems with this code.

  1. You are placing the brackets at the wrong place. Instead of

    int ReturningArray()[]

    it should be

    int* ReturningArray()

  2. You are returning a local variable. Local variables only exist during the execution of the function and will be removed afterwards.

In order to make this work you will have to malloc the int array and return the pointer to the array:

#include <stdio.h>
#include <malloc.h>

#define SIZE 5

int* ReturningArray()
{
    int *myIntArray = (int *)malloc(SIZE * sizeof(int));
    myIntArray[0] = 1;
    myIntArray[1] = 2;
    myIntArray[2] = 3;
    myIntArray[3] = 4;
    myIntArray[4] = 5;
    return myIntArray;
}

int main(void)
{
        int i;
        int* myArray = ReturningArray();
        for(i=0;i<SIZE;i++) { 
                printf("%d\n", myArray[i]); 
        }
        free(myArray); // free the memory again
        system("PAUSE");
        return 0;
}

Upvotes: 2

Mike Kwan
Mike Kwan

Reputation: 24447

You're not returning an int, but you're returning the array. This is the same value as &myIntArray[0]. int ReturningArray()[] is not a valid function prototype.

Upvotes: 4

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