Calling Destructor & Virtual table c++

IF you really know how virtual functions work,it is really fun. Here is such an example(for fun i wrote a function fun  )

class base

{

public :

   virtual void fun()

   {

        cout <"base fun";

   }

  ~base(){}

};

class deriv: public base

{

    void fun()

    {

         cout << "deriv fun";

    }

   ~ deriv(){}

};

very simple two classes..

We know virtual funtions are initlaized in constructor and removed to their old base versions in the destructor.

Here is One more proof .

Try these if you are interested , otherwise go to www.youtube.com

base *p = new deiv();

p->fun();
p->~base();
p->fun();

the outputs will be first

deriv fun and

base fun

Because when we call p->~base() it will replaces the virtual pointer with the base version of fun. Thats why the base::fun get called.

We can't call constructor explicity as we did with destructor.

 Ok , I am stopping now.