c - 结构具有 union 时的意外大小

Question

我正在尝试验证结构的大小。由于某些原因，它给了我 18 个字节而不是预期的 14 个字节( union 应该有最大 8 + 2 + 2 = 12 个字节)。有人可以帮助我吗？

typedef struct __attribute__((__packed__)) def
{
        union {
                double x;   /* 8 bytes */
                char *y;    /* 8 bytes as for 64-bit system */
                struct {
                        struct def *array; /* 8 bytes */
                        unsigned short a;  /* 2 bytes */
                        unsigned short b;  /* 2 bytes */
                } z;
        } w;
        unsigned short v;  /* 2 bytes */
} DEF, *DEFP;

int main(int argc, char **argv) {
        printf("size of DEF = %lu\n", sizeof(DEF));
        printf("size of unsigned short = %lu\n", sizeof(unsigned short));
        printf("size of struct def * = %lu\n", sizeof(struct def *));
        printf("size of char * = %lu\n", sizeof(char *));
        printf("size of double = %lu\n", sizeof(double));
}

这是我运行时显示的内容:

$ gcc test.c && ./a.out
size of DEF = 18
size of unsigned short = 2
size of struct def * = 8
size of char * = 8
size of double = 8

Answer 1

__attribute__((__packed__)) 仅指 struct def。它不会将匿名 struct 打包到 struct def 中的匿名 union 中。

很可能是那两个成员

                   unsigned short a;  /* 2 bytes */
                   unsigned short b;  /* 2 bytes */

“使用”4 个但 2 个字节。

---

与您的问题无关:C 要求使用 z 长度修饰符 打印 size_t:

 printf("size of DEF = %zu\n", sizeof (DEF));