是否存在默认静态的成员函数？

Question

Is an overloaded operator new considered static by default? For example:

template <typename E>
class Link
{
private:
    Link<E>* freeList;
public:
    E element;
    Link<E>* next;
    Link(const E& elemVal, Link<E>* nextVal) { element = elemVal; next = nextVal; }
    Link(Link<E>* nextVal) { next = nextVal; }

    void* operator new(size_t)
    {
        Link<E>* temp=freeList;
        freeList=freeList->next;
        return temp;
    }
};

When I tried to compile it, I got the following error:

Invalid use of member 'Link<E>::freeList' in static member function.

I wonder if the overloaded operator new is actually static.

Answer 1

Yes! The overloads for class specific operator new and operator delete are static member functions. They can't be "regular" member functions since there is no instance available at the point of calling them. They are there to (de)allocate raw storage. There is no object instance at that point, for either of them.

This is described in [class.free]/1 :

Any allocation function for a class T is a static member (even if not explicitly declared static).

And [class.free]/5 :

Any deallocation function for a class X is a static member (even if not explicitly declared static).

Answer 2

Yes, the static keyword is optional for operator new , when it's defined as member function of a class it'll always be static member function.

(emphasis mine)

Both single-object and array allocation functions may be defined as public static member functions of a class (versions (15-18)). If defined, these allocation functions are called by new-expressions to allocate memory for single objects and arrays of this class, unless the new expression used the form ::new which bypasses class-scope lookup. The keyword static is optional for these functions: whether used or not, the allocation function is a static member function.