Question

Is allocating a buffer via new char[sizeof(T)] guaranteed to allocate memory which is properly aligned for the type T, where all members of T has their natural, implementation defined, alignment (that is, you have not used the alignas keyword to modify their alignment).

I have seen this guarantee made in a few answers around here but I’m not entirely clear how the standard arrives at this guarantee. 5.3.4-10 of the standard gives the basic requirement: essentially new char[] must be aligned to max_align_t.

What I’m missing is the bit which says alignof(T) will always be a valid alignment with a maximum value of max_align_t. I mean, it seems obvious, but must the resulting alignment of a structure be at most max_align_t? Even point 3.11-3 says extended alignments may be supported, so may the compiler decide on its own a class is an over-aligned type?

Answer 1

What I’m missing is the bit which says alignof(T) will always be a valid alignment with a maximum value of max_align_t. I mean, it seems obvious, but must the resulting alignment of a structure be at most max_align_t ? Even point 3.11-3 says extended alignments may be supported, so may the compiler decide on its own a class is an over-aligned type ?

As noted by Mankarse, the best quote I could get is from [basic.align]/3:

A type having an extended alignment requirement is an over-aligned type. [ Note:
every over-aligned type is or contains a class type to which extended alignment applies (possibly through a non-static data member). —end note ]

which seems to imply that extended alignment must be explicitly required (and then propagates) but cannot

I would have prefer a clearer mention; the intent is obvious for a compiler-writer, and any other behavior would be insane, still…