sizeof的工会大于预期。如何式排列发生在这里？ [英] sizeof union larger than expected. how does type alignment take place here?

问题描述

#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;
}

结果是：

$ 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

为什么加入4字节工会U2的嵌套结构S1的整8个字节增加了工会作为一个整体的尺寸？

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?

推荐答案

该结构 u2.s2 是因为对齐约束16个字节。编译器保证，如果你做出这样结构的数组，每个指针将在8字节边界上对齐。字段 *我需要8个字节，那么Ĵ需要4个字节，编译器插入4字节的填充。因为结构是16个字节，包含它的工会也是16字节。

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.

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