为什么sizeof（a）为16？ （sizeof int是4）

Question

#include <stdio.h>

int main() {
    int *a[2]; // an array of 2 int pointers
    int (*b)[2];
    // pointer to an array of 2 int (invalid until assigned) //
    int c[2] = {1, 2}; // like b, but statically allocated

    printf("size of int %ld\n", sizeof(int));
    printf("size of array of 2 (int *) a=%ld\n", sizeof(a));
    printf("size of ptr to an array of 2 (int) b=%ld\n", sizeof(b));
    printf("size of array of 2 (int) c=%ld\n", sizeof(c));
    return 0;
}

a is an array of 2 integer pointers, so shouldn't the size be 2 * 4 = 8 ?

Tested on GCC.

Answer 1

You're probably compiling on a 64-bit machine where pointers are 8 bytes.

int *a[2] is an array of 2 pointers. Therefore sizeof(a) is returning 16.
(So it has nothing to do with the size of an int .)

---

If you compiled this for 32-bit, you'll mostly get sizeof(a) == 8 instead.

Answer 2

On 64-bit machines, pointers are usually 8 bytes. So the size of an array of two pointers is usually 16 bytes.

int *a[2];    // array of two pointers to int
int (*b)[2];  // pointer to an array of two int

sizeof a;     // is 2 * 8 bytes on most 64-bit machines
sizeof b;     // is 1 * 8 bytes on most 64-bit machines