Why would you make a destructor virtual?
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Answer / chandra
Vitual destructor is used
(1) whenever the base class object pointer points to
derived class object,and is being created using new.
i.e class base - base class
in main - base bptr;
class derived - derived class
in main - bptr = new derived();
Then whenever we use "delete bptr" at runtime always the
memory of bptr is freed but not derived because here the
pointer is of type base and not of derived.
Note: The destructor that gets invoked is the one that
associated with the type of the object.
Simple rule of thumb: Make your destructor virtual if your
class has any virtual functions.
Example:
#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;}
};
void main()
{
Base *Var = new Derived();
delete Var;
}
| Is This Answer Correct ? | 9 Yes | 0 No |
Answer / chandra sekhar
what chandra said is correct.but it is not 100% correct.for
that answer i want to add one point.
whenever
1)there is a pointer in the base class and we allocated
with the help of new in base class and anthoer pointer in
derived class and agin we allocated with the help of new.
2)whenever the base class object pointer points to
derived class object,and is being created using new.
i.e class base - base class
in main - base bptr;
class derived - derived class
in main - bptr = new derived();
this will be illustrated with an example given below
#include <iostream.h>
class Base
{ protected:
int *ptr;
public:
Base()
{ ptr = new int[10];
cout<<"Constructor: Base"<<endl;
}
virtual ~Base()
{
cout<<"Destructor : Base"<<endl;
delete []ptr;
}
};
class Derived: public Base
{
protected:
int *ptr1;
public:
Derived()
{
ptr1=new int[10];
cout<<"Constructor: Derived"<<endl;
}
~Derived()
{
delete []ptr1;
cout<<"Destructor : Derived"<<endl;
}
};
void main()
{
Base *Var = new Derived;
delete Var;
}
if we are not placing virtual in base class, only base
class destructor will be called because "var is of type
Base".so,
if we are not placing virtual compiler will see only "type
of the pointer,not the object which it is holding".so,if we
place virtual it will see the object it is holding.
now memory allocated for ptr,ptr1 in heap was destroyed by
destructors of the respective class.
| Is This Answer Correct ? | 4 Yes | 0 No |
Answer / sunny bogawat
we make destructure as virtual because when we derived a
class from base class then destructor called from derived
to base so for this happening correctly we haveto make base
class destructor as avirtual.
| Is This Answer Correct ? | 0 Yes | 0 No |
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