Access to method pointer to protected method?

Question

This code:

class B {
 protected:
  void Foo(){}
}

class D : public B {
 public:
  void Baz() {
    Foo();
  }
  void Bar() {
    printf(\"%x\\n\", &B::F         


        

Answer 1

Is there a way to mark something fully accessible from derived classes rather than only accessible from derived classes and in relation to said derived class?

Yes, with the passkey idiom. :)

class derived_key
{
    // Both private.
    friend class derived;

    derived_key() {}
};

class base
{
public:
    void foo(derived_key) {}
};

class derived : public base
{
public:
    void bar() { foo(derived_key()); }
};

Since only derived has access to the contructor of derived_key, only that class can call the foo method, even though it's public.
The obvious problem with that approach is the fact, that you need to friend every possible derived class, which is pretty error prone. Another possible (and imho better way in your case) is to friend the base class and expose a protected get_key method.

class base_key
{
    friend class base;

    base_key() {}
};

class base
{
public:
    void foo(base_key) {}

protected:
    base_key get_key() const { return base_key(); }
};

class derived1 : public base
{
public:
    void bar() { foo(get_key()); }
};

class derived2 : public base
{
public:
    void baz() { foo(get_key()); }
};

int main()
{
  derived1 d1;
  d1.bar(); // works
  d1.foo(base_key()); // error: base_key ctor inaccessible
  d1.foo(d1.get_key()); // error: get_key inaccessible

  derived2 d2;
  d2.baz(); // works again
}

See the full example on Ideone.