Name lookup differences between g++ and MSVS

Question

Consider this code:

#include <iostream>

namespace N {
    class A {};
    void f(A a) { std::cout << "N::f\n"; }
}

void f(int i) { std::cout << "::f\n"; }

template <typename T>
class Base {
  public:
    void f(T x) { std::cout << "Base::f\n"; }
};


template <typename T>
class X : public Base<T> {
  public:
    void g() {
        T t;
        f(t);
    }
};

int main()
{
    X<N::A> x1;
    x1.g();

    X<int> x2;
    x2.g();
}

The code is intended to investigate how name lookup works in C++.

If I compile this program with GNU C++ (version 6.1.0), it prints:

N::f
::f

But if I compile it with Microsoft Visual Studio 2015, it prints:

Base::f
Base::f

Which is the correct behaviour, and why?

Answer 1

g++ is Standard compliant here, and Visual C++ not:

14.6.2 Dependent names [temp.dep]

3 In the definition of a class or class template, the scope of a dependent base class (14.6.2.1) is not examined during unqualified name lookup either at the point of definition of the class template or member or during an instantiation of the class template or member.

Replacing f() with this->f() will find the base member.

Answer 2

In the function definition of the function g the name f is considered as a function declared outside the class (within the class definition this name is not declared; f is a dependent name).

template <typename T>
class X : public Base<T> {
  public:
    void g() {
        T t;
        f(t);
    }
};

So the compiler uses the ADL lookup.

However if to write an explicit call of a member function

class X : public Base<T> {
  public:
    void g() {
        T t;
        this->f(t);
    }
};

then the call of the function f will be considered as a call of the member function of the base class..

Thus it seems that MS VC++ 2015 has a bug.