Private function member called outside of class

Question

In the example below, why is B::f() called even though it is private?

I know this fact that : Access is checked at the call point using the type of the expression used to denote the object for which the member function is called.

#include <iostream>

class A {
public:
  virtual void f() { std::cout << "virtual_function"; }
};

class B : public A {
private:
  void f() { std::cout << "private_function"; }
};

void C(A &g) { g.f(); }

int main() {
  B b;
  C(b);
}

Answer 1

Because the standard says so:

[C++11: 11.5/1]: The access rules (Clause 11) for a virtual function are determined by its declaration and are not affected by the rules for a function that later overrides it. [ Example:

class B {
public:
  virtual int f();
};
class D : public B {
private:
  int f();
};
void f() {
  D d;
  B* pb = &d;
  D* pd = &d;
  pb->f();       // OK: B::f() is public,
                 // D::f() is invoked
  pd->f();       // error: D::f() is private
}

—end example ]

The example is the same as yours, lol.

Answer 2

private functions can override public virtual functions from a base class. Accessability is, in a fact, completely ignored when determining whether a function overrides another, so even in

// Redundant private for clarity:
class A { private: virtual void foo(); };
class B : A { public: void foo(); };

B::foo overrides A::foo.