この仕組みを教えて下さい
もしかして2bit+1byte=2bytesってこと?

> > Option<Option<T>> has layout [0..1][0..1]<u8> , i.e., be of size 3
>
> False. Thanks to @eddyb’s work, the compiler will collapse the discriminant of the first option into the second. Thus, mem::size_of::<Option<Option<u8>>>() == 2.