How do virtual destructors work?

Question

I am using gcc. I am aware how the virtual destructors solve the problem when we destroy a derived class object pointed by a base class pointer. I want to know how do they work?

class A
{
      public:
      A(){cout<<"A constructor"<<endl;}
     ~A(){cout<<"A destructor"<<endl;}

};

class B:public A
{
      public:
      B(){cout<<"B constructor"<<endl;}
      ~B(){cout<<"B destructor"<<endl;}
};       

int main()
{
  A * a = new B();
  delete a;    
  getch();
  return 0;   
} 

When I change A's destructor to a virtual function, the problem is solved. What is the inner working for this. Why do I make A's destructor virtual. I want to know what happens to the vtable of A and B?

Answer 1

Virtual destructor is just a virtual function, so it adheres to the same rules.

When you call delete a, a destructor is implicitly called. If the destructor is not virtual, you get called a->~A(), because it's called as every other non-virtual function.

However if the destructor is virtual, you get ~B() called, as expected: the destructor function is virtual, so what gets called is the destructor of derived class, not base class.

Edit:
Note that the destructor of the base class will be called implicitly after the destructor of the derived class finishes. This is a difference to the usual virtual functions.

Answer 2

The key thing you need to know is that not using a virtual destructor in the above code is undefined behavior and that's not what you want. Virtual destructors are like any other virtual functions - when you call delete the program will decide what destructor to call right in runtime and that solves your problem.