压缩结构的大小，在C中具有小于8位的位字段的并集

Question

Is it possible in C to get the size of the following structure to be 2?

#include <stdio.h>

struct union_struct {
    char foo;
    char bar : 2;
    union {
        char foobar1 : 6;
        char foobar2 : 6;
    };
};

int main(void)
{
    printf("size of union struct: %d\n", sizeof(struct union_struct));
    return 0;
}

output, compiled with gcc:

size of union struct: 3

Answer 1

If you are relying on implementation defined behavior, then yes, but you have to organize it a bit differently:

#ifdef UNNAMED_BITFIELDS_ARE_WELL_DEFINED
#define ANON
#else
#define ANON3(X) anonymous__## X ##__
#define ANON2(X) ANON3(X)
#define ANON ANON2(__LINE__)
#endif

struct union_struct {
    char foo;
    union {
        struct {
            char bar     : 2;
            char ANON    : 6;
        };
        struct {
            char ANON    : 2;
            char foobar1 : 6;
        };
        struct {
            char ANON    : 2;
            char foobar2 : 6;
        };
    };
};

The first byte is foo , the second byte is the anonymous union. Then anonymous union has 3 single byte anonymous structs. Each struct (optionally) uses unnamed bit-fields to allow foobar1 and foobar2 to represent the same 6 bits that follow bar .

From my understanding of the C.11 standard the above code is correct when UNNAMED_BITFIELDS_ARE_WELL_DEFINED is defined. However, there seems to be debate on whether or not unnamed bit-fields have well-defined semantics (see comments below). If unnamed bit-fields do not have well-defined semantics, then the code above can expand each ANON macro into a name for the bit-field.

However, the C.11 standard only defines bit-fields on _Bool , int , and unsigned , while the use of any other type for a bit-field is implementation defined (C.11 §6.7.2.1 ¶5).