Placement new in std::aligned_storage?

Question

Suppose I have a type template parameter T.

And suppose I have a std::aligned_storage as follows:

typename std::aligned_storage<sizeof(T), alignof(T)>::type storage; 

I want to placement new a T into the storage.

What is the standard-compliant pointer value/type to pass to the placement new operator, and how do I derive that from storage?

new (& ???) T(a,b,c); 

For example:

new (&storage) T(a,b,c); new (static_cast<void*>(&storage)) T(a,b,c); new (reinterpret_cast<T*>(&storage)) T(a,b,c); new (static_cast<T*>(static_cast<void*>(&storage)); 

Which of the above (if any) are compliant, and if none, what is the better way?

Answer 1

The most paranoid way is

::new ((void *)::std::addressof(storage)) T(a, b, c); 

Explanation:

• ::std::addressof guards against overloaded unary operator& on storage, which is technically allowed by the standard. (Though no sane implementation would do it.) The ::std guards against any non-top-level namespaces (or classes) called std that might be in scope.
• (void *) (which in this case is the equivalent of a static_cast) ensures that you call the placement operator new taking a void * rather than something else like decltype(storage) *.
• ::new skips any class-specific placement operator news, ensuring that the call goes to the global one.

Together, this guarantees that the call goes to the library placement operator new taking a void *, and that the T is constructed at where storage is.

In most sane programs, though,

new (&storage) T(a,b,c); 

should be sufficient.

Answer 2

The placement allocation function is described as follows (C++14 n4140 18.6.1.3):

void* operator new(std::size_t size, void* ptr) noexcept; 

Returns: ptr.

Remarks: Intentionally performs no other action.

20.10.7.6 table 57 describes aligned_storage<Len, Align> thus:

The member typedef type shall be a POD type suitable for use as uninitialized storage for any object whose size is at most Len and whose alignment is a divisor of Align.

This implies that in your case, &storage is suitably aligned for holding an object of type T. Therefore, under normal circumstances¹, all 4 ways you've listed of calling placement new are valid and equivalent. I would use the first one (new (&storage)) for brevity.

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¹ T.C. correctly pointed out in the comments that it is technically possible for your program to declare an overload of the allocation function taking a typename std::aligned_storage<sizeof(T), alignof(T)>::type*, which would then be selected by overload resolution instead of the library-provided 'placement new' version.

I would say this unlikely in at least 99.999% of cases, but if you need to guard against that as well, use one of the casts to void*. The direct static_cast<void*>(&storage) is enough.

Also, if you're paranoid to this level, you should probably use ::new instead of just new to bypass any class-specific allocation functions.