Does delete on a pointer to a subclass call the base class destructor?

Question

I have an class A which uses a heap memory allocation for one of its fields. Class A is instantiated and stored as a pointer field in another class (clas

Answer 1

You should delete A yourself in the destructor of B.

Answer 2

It is named "destructor", not "deconstructor".

Inside the destructor of each class, you have to delete all other member variables that have been allocated with new.

edit: To clarify:

Say you have

struct A {}

class B {
    A *a;
public:
    B () : a (new A) {}
    ~B() { delete a; }
};

class C {
    A *a;
public:
    C () : a (new A) {}        
};

int main () {
    delete new B;
    delete new C;
}

Allocating an instance of B and then deleting is clean, because what B allocates internally will also be deleted in the destructor.

But instances of class C will leak memory, because it allocates an instance of A which it does not release (in this case C does not even have a destructor).

Answer 3

If you have a usual pointer (A*) then the destructor will not be called (and memory for A instance will not be freed either) unless you do delete explicitly in B's destructor. If you want automatic destruction look at smart pointers like auto_ptr.

Answer 4

No. the pointer will be deleted. You should call the delete on A explicit in the destructor of B.

Answer 5

no it will not call destructor for class A, you should call it explicitly (like PoweRoy told), delete line 'delete ptr;' in example to compare ...

  #include <iostream>

  class A
  {
     public:
        A(){};
        ~A();
  };

  A::~A()
  {
     std::cout << "Destructor of A" << std::endl;
  }

  class B
  {
     public:
        B(){ptr = new A();};
        ~B();
     private:
        A* ptr;
  };

  B::~B()
  {
     delete ptr;
     std::cout << "Destructor of B" << std::endl;
  }

  int main()
  {
     B* b = new B();
     delete b;
     return 0;
  }

Answer 6

class B
{
public:
    B()
    {
       p = new int[1024];  
    }
    virtual ~B()
    {
        cout<<"B destructor"<<endl;
        //p will not be deleted EVER unless you do it manually.
    }
    int *p;
};


class D : public B
{
public:
    virtual ~D()
    {
        cout<<"D destructor"<<endl;
    }
};

When you do:

B *pD = new D();
delete pD;

The destructor will be called only if your base class has the virtual keyword.

Then if you did not have a virtual destructor only ~B() would be called. But since you have a virtual destructor, first ~D() will be called, then ~B().

No members of B or D allocated on the heap will be deallocated unless you explicitly delete them. And deleting them will call their destructor as well.