C Bit-Field Pitfalls

Therefore, convert the values before operating on them:

    u1 = ((uint64_t) bf.uf1) << 20;
    u2 = ((uint64_t) bf.uf2) << 20;

> Each of the comma-separated multisets designates the same type, except that for bit-fields, it is implementation-defined whether the specifier `int` designates the same type as `signed int` or the same type as `unsigned int`.

That means a bit-field member using plain `int` as the underlying type might itself be signed or unsigned, similar to whether plain `char` is signed or unsigned. There's a proposal to address this: https://www.open-std.org/jtc1/sc22/wg14/www/docs/n3828.pdf Fortunately, it found that GCC, Clang, MSVC, and ICC all treat `int` bit-fields as `signed int`, and the recommendation is to require this behavior.

That said, I don't think I've ever seen a bit-field deliberately used for signed values; there's some logic to the allowance granted by the standard. And I'm sure there's plenty of real-world code erroneously using `int` bit-fields for unsigned values that just happens to work because of twos-complement representations and the semantics of bitwise operations. But better to limit this kind of flexibility, especially when real-world implementations seem not to have taken the alternative route.

I suppose you could say MS's choice reflects a commitment to backwards compatibility, whereas GCC/Clang is always chomping at the bit to introduce more aggressive optimizations that signed-integer-undefined-behavior affords?

this is nonsense. I don't know what they expect would happen, but who cares? I wouldn't shift a 12 bit field by more than ±11 bits.

you can shift the "enclosing" word of memory if you want, just put the original definition in a union.