Is sizeof(T) == sizeof(int)?

Question

I\'ve been poring over the draft standard and can\'t seem to find what I\'m looking for.

If I have a standard-layout type

struct T {
   unsigned hand         


        

Answer 1

You essentially are asking, given:

struct T {
    U handle;
};

whether it's guaranteed that sizeof(T) == sizeof(U). No, it is not.

Section 9.2/17 of the ISO C++03 standard says:

A pointer to a POD-struct object, suitably converted using a reinterpret_cast, points to its initial member (or if that member is a bit-field, then to the unit in which it resides) and vice versa.

Suppose you have an array of struct T. The vice versa part means that the address of any of the T::handle members must also be a valid address of a struct T. Now, suppose that these members are of type char and that your claim is true. This would mean that struct T would be allowed to have an unaligned address, which seems rather unlikely. The standard usually tries to not tie the hands of implementations in such a way. For your claim to be true, the standard would have to require that struct T be allowed to have unaligned addresses. And it would have to be allowed for all structures, because struct T could be a forward-declared, opaque type.

Furthermore, section 9.2/17 goes on to state:

[Note: There might therefore be unnamed padding within a POD-struct object, but not at its beginning, as necessary to achieve appropriate alignment.]

Which, taken a different way, means that there is no guarantee that there will never be padding.