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

一旦A的析构函数被声明为virtual ，所有派生类的析构函数也是virtual ，即使它们没有明确声明为这样。所以你看到的行为正是预期的

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();

REFERENCE

Answer 3

As @juanchopanza said - declaring the base destructor virtual means all descendants have virtual destructors. This inherited virtuality is the same for any methods , not just destructors.

It's why I have interviewed people who didn't know what the keyword did because they had only ever had to override methods derived from a framework, so they were all virtual (sigh).