Question

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

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

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

int main()
{

   A *pointA = new A;
   A *pointB = new B;
   A *pointC = new C;

   delete pointA;
   delete pointB;
   delete pointC;
}

Answer 1

It will invoke undefined behavior at the second (and third) delete, because A’s destructor is not virtual.

§5.3.5/3:

if the static type of the operand is different from its dynamic type, the
static type shall be a base class of the operand’s dynamic type and the static type shall have a virtual destructor or the behavior is undefined.

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If you make the destructor of A virtual, you get well-defined behavior, and the destructor of the dynamic type is called. (And each of those in turn calls the base destructor.) Your output would be:

A destructor
B destructor
A destructor
C destructor
B destructor
A destructor

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For what it’s worth, when you’re that close to a compilable snippet, you should leave the includes. Also, just use struct instead of class to be concise about the public stuff, and leave out empty constructors.