sizeof Explanation

[rachitagrawal]

Can someone please explain the behviour of this code

#include <stdio.h>
#define SIZE 10

void size(int arr[SIZE])
{
    int b[10];
    printf("size of array is:%d\n",sizeof(arr));
    printf("size of array is:%d\n",sizeof(b));
}

int main()
{
    int arr[SIZE];
    size(arr);
    return 0;
}

[chrisname]

sizeof returns the size of a variable (bytes, e.g. sizeof(int) on most PCs will be 4; sizeof(char) will be 1 on most PCs, etc). In this case it is returning the amount of elements in the arrays, so ten.

[helios]

In this case it is returning the amount of elements in the arrays, so ten.

You sure about that?

[Duthomhas]

The sizeof operator always returns the number of bytes used by its argument, using the most complete information available.

The int arr[WHATEVER] is the same as an int*, which has the size of a pointer to int.

However, the 'b' on line 6 is a known array type, so its size is the same as the size of ten ints.

Keep in mind that the argument to your 'size' function on line 4 is not necessarily the 'arr' aray declared on line 13. It makes no difference that they are declared with the same name and subscripts -- they are, quite simply, not necessarily the same objects. Hence, no further information about the original object is available inside the function (lines 5 through 9).

When used as a pointer, an array will degrade its type to a pointer to the first element. The reverse is not possible.

Hope this helps.

[chrisname]

You sure about that?

I was fairly sure it would return the amount of elements in an array. It usually does, anyway... Theoretically it should return the same value as (strlen(arr)) * int* but it never seems to.

[helios]

(strlen(arr)) * int*

Huh??

it never seems to

Have you considered that maybe that's because your theory is wrong?

[chrisname]

Have you considered that maybe that's because your theory is wrong?

No. I'm never wrong.

On a serious note; yes. But it always seems to return the amount of elements in an array.

http://en.wikipedia.org/wiki/Sizeof#Using_sizeof_with_arrays

When sizeof is applied to an array, the result is the size in bytes of the array in memory.

(strlen(arr)) * int*

Sorry, I meant (strlen(arr)) * sizeof(int*).

[firedraco]

When sizeof is applied to an array

^That is the key thing. In the function, you pass a pointer to the array, not the array itself.

[helios]

Sorry, I meant (strlen(arr)) * sizeof(int*).

I must have unconsciously compensated for your error, because I understood it like that. It still doesn't make sense.