推导类型的基类成员的指针

Question

#include <type_traits>

struct BaseClass
{
    int baseValue;
};

struct DerivedClass : public BaseClass
{
    int derivedValue;
};

int main()
{
    auto memDataPtr = &DerivedClass::baseValue;

    static_assert(std::is_same<decltype(memDataPtr), int BaseClass::*>::value, "Huh?");
}

All the compilers I've tried this on compile this code successfully, ie, the static assert does not fire. This means that if I take the address of a member inherited from a base class, the resultant pointer-to-member refers to the base class the member comes from, even if I explicitly use the derived class name when specifying the member.

I'm curious to know where in the spec this is explicitly called out, and the reasoning behind this behavior, as it seems a bit counter-intuitive (to my naive self at least).

Answer 1

See [expr.unary.op], which says

If the operand is a qualified-id naming a non-static or variant member m of some class C with type T, the result has type “pointer to member of class C of type T” and is a prvalue designating C::m.

It also gives an example to show this behavior:

struct A { int i; };
struct B : A { };
... &B::i ... // has type int A::*