What does the scope resolution operator return in case of Type::var?

Question

Consider the following example:

class A { int x; };

Now what is A::x?

• It cannot be an lvalue because it does not refer to a storage Location.
• It cannot be a type, because the type would be decltype(A::x).

Answer 1

It is, in fact, an lvalue. [expr.prim.id.qual]/2:

A nested-name-specifier that denotes a class [...] followed by the name of a member of either that class ([class.mem]) or one of its base classes, is a qualified-id [...]. The result is the member. The type of the result is the type of the member. The result is an lvalue if the member is a static member function or a data member and a prvalue otherwise.

Though its usage outside a class member access expression is severely restricted by [expr.prim.id]/2, it can notably be used in unevaluated operands, where its lvalueness can manifest:

struct A {
    int x;
};

void f(int &);

using p = decltype(&(A::x)); // p is int*; the parens prevents forming a pointer-to-member
using q = decltype(f(A::x)); // q is void