How does the derived class destructor get invoked being private in the following program?

Question

#include<iostream>

class base
{
   public:
       virtual ~base(){std::cout << "base\n";}
};
class derived : public base
{
   private:
        ~derived(){std::cout << "derived\n";} /* destructor is private */
};
int main()
{
      base *pt= new derived;
      delete pt;
}

The above program compiles and runs fine. How does the derived class destructor get invoked being private ?

Answer 1

This will happen not only with destructors.

You can override any virtual public function with a private one.

#include<iostream>

class base
{
   public:
       virtual void x(){std::cout << "base\n";}
};
class derived : public base
{
   private:
        void x(){std::cout << "derived\n"; base::x();}
};
int main()
{
      base *pt= new derived;
      pt->x(); //OK
      //((derived *)pt)->x();  //error: ‘virtual void derived::x()’ is private
      derived *pt2= new derived;
      //pt2->x();  //error: ‘virtual void derived::x()’ is private
      ((base *)pt2)->x(); //OK
}

Upside/downside of this is you will have to use pointer to base to access this method. This feature is one of ways to separate public API from customized implementation.

So, in other words, your destructor is getting called because it was declared as public in base and you are calling it trough pointer to base.