Why std::allocator<T>::allocate calls ::operator new?

Question

The documentation for the std::allocator<T>::allocate member function says in ([allocator.members]) that:

Remarks: The storage for the array is obtained by calling ​::​operator new ([new.delete]), but it is unspecified when or how often this function is called. This function starts the lifetime of the array object, but not that of any of the array elements.

I wonder why it says ::operator new and not just operator new? Does the double colon make any difference? Which other operator new could be called here, if that double colon was omitted?

Elephant

https://mcmap.net/q/757833/-what-are-quot-operator-new-quot-and-quot-operator-delete-quot – Cassondracassoulet

@Cassondracassoulet I can understand that, in general, we sometimes need to distinguish between the global and class-specific operator new. But how is this related to std::allocator? – Elephant

new is an operator, and as such is subject to the rules of ADL like other operators. By using ::operator new the ADL gets bypassed and only the global new defined by the compiler, or the global new you have replaced it with, will be used. – Sklar

@Sklar I understand that. But I don't know which other possible operator new could be selected by ADL here than that global one (compiler-defined or replaced)? – Elephant

Classes can overload operator new. If they did so and allocate just used new T then you would get that class overload of new instead of the global new. ::operator new stops that from happening even if the class overloads new – Sklar

@Sklar How can std::allocator<T>::allocate use new T? It should just allocate and not initialize any objects of type T. Or am I wrong? – Elephant

@DanielLangr Sorry, should have said if they used operator new, and now that I say that I get your confusion. They would have to use T::operator new to get the class new. – Sklar

Answer 1

Prior to LWG2818, [contents]p3 read:

Whenever a name x defined in the standard library is mentioned, the name x is assumed to be fully qualified as ::std::x, unless explicitly described otherwise. For example, if the Effects: element for library function F is described as calling library function G, the function ::std::G is meant.

So writing operator new in the specification would mean ::std::operator new, which wouldn't make sense. ::operator new correctly refers to operator new in the global namespace.

In an implementation of the standard library, there would be no difference between writing operator new and ::operator new since std::allocator<T> wouldn't define a member operator new and ADL has no effect since there is only one namespace a free operator new could be defined in.

Flagship

It is exactly the explanation I was looking for. Thanks! I didn't know about this rule and learnt something new today. – Elephant