Access a derived private member function from a base class pointer to a derived object [duplicate]

Question

Possible Duplicate:
Why can i access a derived private member function via a base class pointer to a derived object?

#include <iostream>
using namespace std;

class B {
public:
  virtual void fn1(void) {cout << "class B : fn  one \n"; }
  virtual void fn2(void) {cout << "class B : fn  two \n"; }
};

class D: public B {
    void fn1(void) {cout << "class D : fn one \n"; }
private:
    void fn2(void) {cout << "class D : fn two \n"; }
};

int main(void)
{
    B *p = new D;

    p->fn1();
    p->fn2();
}

Why does p->fn2() call the derived class function even though fn2 is private in D ?

Answer 1

Access modifiers, such as public, private and protected are only enforced during compilation. When you call the function through a pointer to the base class, the compiler doesn't know that the pointer points to an instance of the derived class. According to the rules the compiler can infer from this expression, this call is valid.

It is usually a semantic error to reduce the visibility of a member in a derived class. Modern programming languages such as Java and C# refuse to compile such code, because a member that is visible in the base class is always accessible in the derived class through a base pointer.