sizeof union larger than expected. How does type alignment take place here?

Question

#include <stdio.h>

union u1 {
    struct {
        int *i;
    } s1;
    struct {
        int i, j;
    } s2;
};

union u2 {
    struct {
        int *i, j;
    } s1;
    struct {
        int i, j;
    } s2;
};

int main(void) {
    printf("        size of int: %zu\n", sizeof(int));
    printf("size of int pointer: %zu\n", sizeof(int *));
    printf("   size of union u1: %zu\n", sizeof(union u1));
    printf("   size of union u2: %zu\n", sizeof(union u2));
    return 0;
}

Results in:

$ gcc -O -Wall -Wextra -pedantic -std=c99 -o test test.c
$ ./test
        size of int: 4
size of int pointer: 8
   size of union u1: 8
   size of union u2: 16

Why does adding an integer of 4 bytes to nested struct s1 of union u2 increase the size of the union as a whole by 8 bytes?

Answer 1

The struct u2.s2 is 16 bytes because of alignment constraints. The compiler is guaranteeing that if you make an array of such structs, each pointer will be aligned on an 8-byte boundary. The field *i takes 8 bytes, then j takes 4 bytes, and the compiler inserts 4 bytes of padding. Because the struct is 16 bytes, the union containing it is also 16 bytes.