c - 如何获取内部结构的大小

Question

我正在尝试获取内部结构的大小，即 struct B。但是我遇到了编译错误:

prog.c: In function ‘main’: prog.c:10:53: error: expected ‘)’ before ‘:’ token printf("%d | %d", sizeof(struct A), sizeof(struct A::struct B));

以下是我的代码:

#include <stdio.h>

struct A
{
        struct B{};
};

int main() {
    printf("%d | %d", sizeof(struct A), sizeof(struct A::struct B));
    return 0;
}

你能建议我如何在 C 中实现这个吗？

已更新

来自 @Jabberwocky 的回答解决了我的上述问题。但是下面的代码呢？这也可以找到here :

#include <stdio.h>

struct A
{
    struct B{};
};

struct B
{};

int main() {
    printf("%d | %d", sizeof(struct A), sizeof(struct B), sizeof(struct A::struct B));
    return 0;
}

在这种情况下，我得到如下编译错误:

prog.c:8:8: error: redefinition of ‘struct B’
struct B
^
prog.c:5:10: note: originally defined here
struct B{};
^
prog.c: In function ‘main’:
prog.c:12:71: error: expected ‘)’ before ‘:’ token
printf("%d | %d", sizeof(struct A), sizeof(struct B), sizeof(struct A::struct B));

这里我如何区分 struct B 和 struct A::struct B

Answer 1

#include <stdio.h>

struct A
{
        struct B{};   // this should not compile anyway in C as C
                      // does not allow empty structs
                      // but this would compile: struct B{int a;};
};

int main() {
    printf("%d | %d", sizeof(struct A), sizeof(struct B));
                                           // just write struct B
    return 0;
}

工作样本:

#include <stdio.h>

struct A
{
  int b;
  struct B { int a; };
};

int main() {
  printf("%d | %d", sizeof(struct A), sizeof(struct B));
  return 0;
}

在 32 位系统上可能的输出:

8 | 4