Why in this program size of integer is 8?

April 3, 2023

// pointer of array

#include<stdio.h>

int main()
{
    int arr[] = { 3, 5, 6, 7, 9 };
    int *p = arr;
    int (*ptr)[5] = &arr;

    printf("p = %p, ptr = %p\n", p, ptr);

    printf("*p = %d, *ptr = %p\n", *p, *ptr);

    printf("sizeof(p) = %lu, sizeof(*p) = %lu\n",
                          sizeof(p), sizeof(*p));

    printf("sizeof(ptr) = %lu, sizeof(*ptr) = %lu\n",
                        sizeof(ptr), sizeof(*ptr));

    return 0;
}

Output

p = 0x7ffde1ee5010, ptr = 0x7ffde1ee5010
*p = 3, *ptr = 0x7ffde1ee5010
sizeof(p) = 8, sizeof(*p) = 4
sizeof(ptr) = 8, sizeof(*ptr) = 20

Here why sizeof(p) or sizeof(ptr) is 8?

This code is an example of pointer to array

Why 8 and why not 4 ?

>Solution :

On a 64-bit system, which yours evidently is, pointers are represented by 8 bytes, regardless of what they point to.

In your example, both p and ptr are pointers. In the case of p, we have a pointer to the first int in the array. On your system evidently int is 4 bytes. In the case of ptr, it’s been declared as a pointer to an array of 5 int values, which is 20 bytes.