c - 为什么 sizeof 未命名位域成员结构打印 1？

Question

在下面的程序中，在结构中声明了未命名的位域成员。

#include <stdio.h>

struct st{
    int : 1;
};

int main()
{
    struct st s;
    printf("%zu\n",sizeof(s));  // print 1
}

以上程序打印输出1。

为什么 sizeof(s) 打印 1？

Answer 1

sizeof(s) 是 undefined因为结构中没有其他命名成员。

C11 6.7.2.1(P8):

The presence of a struct-declaration-list in a struct-or-union-specifier declares a new type, within a translation unit. The struct-declaration-list is a sequence of declarations for the members of the structure or union. If the struct-declaration-list contains no named members, no anonymous structures, and no anonymous unions, the behavior is undefined. The type is incomplete until immediately after the } that terminates the list, and complete thereafter.

如果你这样写:

struct st{
        int : 1;
        int i : 5;
    };

所以，sizeof(s) 是可以的，因为结构中也有命名的位域成员。