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Home/ Questions/Q 8697399
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T01:30:51+00:00 2026-06-13T01:30:51+00:00

Why are all destructors, ~D() , ~C() , ~B() , ~A() being called in

  • 0

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;
}
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  1. Editorial Team
    Editorial Team
    2026-06-13T01:30:52+00:00Added an answer on June 13, 2026 at 1:30 am

    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

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