Is this a standard C++ code?

Question

The following simple piece of code, compiles with VC2008 but g++ rejects the code:

#include <iostream>

class myclass
{
protected:
    void print() { std::cout << "myclass::print();"; }
};

struct access : private myclass
{
    static void access_print(myclass& object)
    {
        // g++ and Comeau reject this line but not VC++
        void (myclass::*function) () = &myclass::print;

        (object.*function)();
    }
};

int main()
{
    myclass object;
    access::access_print(object);
}

(/W4) is turned on in VC, but it doesn't give any warning.

g++ 4.4.1 gives me an error:

correct.cpp: In static member function ‘static void access::access_print(myclass&)’:
correct.cpp:6: error: ‘void myclass::print()’ is protected

If g++ is correct, how do I access a protected member of a class? is there another way?

---

@Suroot Do you mean that I shouldn't pass an object of type myclass? It doesn't matter actually, g++ gives the same error but VC compiles the code without any warning.

#include <iostream>

class myclass
{
protected:
    void print() { std::cout << "myclass::print();"; }
};

struct access : private myclass
{
    static void access_print()
    {
        myclass object;
        void (myclass::*function) () = &myclass::print;

        (object.*function)();
    }
};

int main()
{
    access::access_print();
}

Answer 1

I believe g++ and comeau are correct. The specifier for a protected member must be of type "access" or derived, so I believe the code:

void (myclass::*function) () = &access::print;

would compile.

I believe this is because of 11.5.1:

... If the access [ed. to a protected member ] is to form a pointer to member, the nested-name-specifier shall name the derived class (or any class derived from that class).