Does new char actually guarantee aligned memory for a class type?

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

The expressions new char[N] and new unsigned char[N] are guaranteed to return memory sufficiently aligned for any object. See §5.3.4/10 "[...] For arrays of char and unsigned char, the difference between the result of the new-expression and the address returned by the allocation function shall be an integral multiple of the strictest fundamental alignment requirement (3.11) of any object type whose size is no greater than the size of the array being created. [ Note: Because allocation functions are assumed to return pointers to storage that is appropriately aligned for objects of any type with fundamental alignment, this constraint on array allocation overhead permits the common idiom of allocating character arrays into which objects of other types will later be placed. —end note ]".

From a stylistic point of view, of course: if what you want is to allocate raw memory, it's clearer to say so: operator new(N). Conceptually, new char[N] creates N char; operator new(N) allocates N bytes.