Anonymous struct padding

Code:

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#include <stdio.h>

struct A {
    int i;
    char a, b, c;
};

struct B {
    struct {
        int i;
        char a, b, c;
    };

    char d;
};

struct C {
    int i;
    char a, b, c, d;
};

int main() {
    printf("sizeof(struct A) = %lu\n", sizeof(struct A));
    printf("sizeof(struct B) = %lu\n", sizeof(struct B));
    printf("sizeof(struct C) = %lu\n", sizeof(struct C));

    return 0;
}

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Compiled and executed with both GCC and Clang. I expect that the `sizeof` structs A, B, and C are all exactly 8 bytes (on a 32 bits system). But `struct B` is 12 bytes. I know why it is 12 bytes. What I don't understand is why the anonymous struct in `struct B` is padded to 8 bytes. Surely the compiler can figure out that `struct B` is already aligned to 4 bytes? What's going on here?

struct A has to be padded to 8 bytes so that it is correctly aligned when you set up an array of them. struct B contains an identical struct to A, plus anohter member. So it's 8 bytes, plus one byte padded to 4 bytes, = 12 bytes.

The compiler isn't clever enough to realise that since the member struct of struct B is anonymous, it can never appear in any other context than struct B, abd that it would be possible to reduce padding a bit more aggressively. It's a rarely-encountered situation, and it would have to special-case it to make this optimisation work.

Thanks, Malcolm Mclean. It is a pity that this is merely an unimplemented optimization. Well, at least in these cases #pragma pack(1) can save me some bytes with no performance cost.

Quote:

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Originally Posted by CoiledAlizarine

Thanks, Malcolm Mclean. It is a pity that this is merely an unimplemented optimization. Well, at least in these cases #pragma pack(1) can save me some bytes with no performance cost.

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Reading/Writing unaligned data always has a performance penalties (specially on Intel/AMD processors).