Using late binding to call a private method from a non private context

Question

I was surprised by the output produced by the piece of code below (g++ 4.4.7).

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

class B : public A {
private:
    // Automatically virtual, because of base class
    void f() {std::cout << "B::f()" << std::endl;}
};

int main(int argc, const char *argv[])
{
    A *pB = new B();
    pB->f();
    return 0;
}

The output is

B::f()

I know that because of late binding the compiler cannot issue an error here, but why can we call a private method from a non-private context?

What is the rationale?

score 2 · Accepted Answer · answered Oct 18 '13 at 11:43

2

n3376 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.

11.5.2

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. The access of the member function in the class in which it was defined (D in the example above) is in general not known.

answered Oct 18 '13 at 11:43

ForEveR

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Exactly the answer I was looking for! Couldn't find it in the standard myself. Thanks. – Vincenzo Pii Oct 19 '13 at 11:29

score 1 · Answer 2 · edited Oct 18 '13 at 11:56

1

Access specifiers only exist for compilation purposes. Any memory within your program's allocation can be accessed by any part of the executable; there is no public/private concept at runtime

Late binding functions are run time so there is no public and private concept for run time

edited Oct 18 '13 at 11:56

Vincenzo Pii

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answered Oct 18 '13 at 11:50

Birendra Kumar

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Can you please indicate the source of the quote (or remove quoting block if those are your own words :))? – Vincenzo Pii Oct 18 '13 at 11:56

score 0 · Answer 3 · answered Oct 18 '13 at 12:11

Actually I have tried to execute this code with little modification to make specify base
class virtual function in private section and I got error from compiler. because in compile
time first control goes to base class function but it is defined in private section so
compiler throw error . Whenever you declare a object of the derived class taking the
reference of base class. In compile time control goes to base class first and every class makes its own virtual table and then run time the pointer object invoke derived class function and no matters where it is declared private or public as I said earlier.

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

class B : public A {
   private:
     // Automatically virtual, because of base class
      void f() {std::cout << "B::f()" << std::endl;}
 };

int main(int argc, const char *argv[])
 {
  A *pB = new B();
   pB->f();
 return 0;
}

We are calling f() through the base class pointer,thats why the control goes to the base class first. — Subhajit, Oct 18 '13 at 13:02

Using late binding to call a private method from a non private context

3 Answers3