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Virtual destructors behave the same way as virtual functions. Consider the code given below:
- Anu December 11, 2010#include <iostream.h>
class Base
{
public:
Base(){ cout<<"Constructor: Base"<<endl;}
virtual ~Base(){ cout<<"Destructor : Base"<<endl;}
};
class Derived: public Base
{
//Doing a lot of jobs by extending the functionality
public:
Derived(){ cout<<"Constructor: Derived"<<endl;}
~Derived(){ cout<<"Destructor : Derived"<<endl;}
};
int main()
{
Base *Var = new Derived();
delete Var;
return 0;
}
Output:
Constructor: Base
Constructor: Derived
Destructor: Derived
Destructor: Base
If the destructor in the base class is not declared virtual, then only the destructor of the base class will be called, as the pointer is of type base. i.e, it won't consider the fact that the pointer actually points to a derived class object.