Why is base-class destructor called on derived object when destructor of derived class is non-virtual?

Question

Why are all destructors, ~D(),~C(),~B(),~A() being called in the example below?

There is only one virtual destructor: that of A.

Here is the code:

#include<iostream>
using namespace std;

class A
{
public:
  virtual ~A()
  {
    cout<<"destruct A\n";
  }

};
class B:public A
{
public:
  ~B()
  {
  cout<<"destruct B\n"; 
  }
};
class C:public B
{
public:
  ~C()
  {
    cout<<"destruct C\n";
  }
};
class D:public C
{
public:
   ~D()
   {
     cout<<"destruct D\n"; 
   }
};

int main()
{
    A* ptr = new D();
    delete ptr;
    return 0;
}

Answer 1

Once A's destructor is declared virtual, the destructors of all derived classes are also virtual, even if they aren't explicitly declared as such.. So the behaviour you see is exactly what is expected

Answer 2

The destruction order in derived objects goes in exactly the reverse order of construction: first the destructors of the most derived classes are called and then the destructor of the base classes.

A destructor can be defined as virtual or even pure virtual. You would use a virtual destructor if you ever expect a derived class to be destroyed through a pointer to the base class. This will ensure that the destructor of the most derived classes will get called:

A* b1 = new B;//if A has a virtual destructor
delete b1;//invokes B's destructor and then A's

A* b1 = new B;//if A has no virtual destructor
    delete b1;//invokes A's destructor ONLY

If A does not have a virtual destructor, deleting b1 through a pointer of type A will merely invoke A's destructor. To enforce the calling of B's destructor in this case we must have specified A's destructor as virtual:

virtual ~A();

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