Dynamic memory allocation with aligned storage

Question

I'm a bit confused about the alignement requirements when std::aligned_storage is used and dynamic allocation is needed. Example:

  using storage = typename std::aligned_storage<sizeof(T), std::alignment_of<T>::value>::type;
  storage* pool;
  pool = ::new storage[num_obj];

Is it legal? The new operator doesn't return alignment memory, so is the result correct?

Answer 1

The C++11 standard requires that allocation functions such as ::operator new return memory that is aligned to alignof(std::max_align_t) [basic.stc.dynamic/2]:

The pointer returned shall be suitably aligned so that it can be converted to a pointer of any complete object type with a fundamental alignment requirement […]

Thus, as long as the type of the object you're creating via your new expression is not an overaligned type (one that requires an alignment more strict than alignof(std::max_align_t)), everything's fine. For overaligned types, you would indeed have to allocate storage of sufficient size to manually align the pointer, e.g., using std::align, and then construct your object at a suitable address, e.g., via placement new…

Starting with C++17, new automatically takes care of this. To allocate storage that requires alignment more strict than __STDCPP_­DEFAULT_­NEW_­ALIGNMENT__ (the alignment that allocation functions are at least required to supply), a new-expression will call an allocation function that is explicitly given the alignment of the storage to be allocated as an argument [expr.new]/14 …