Answer:
If the destructor in the base class is not made virtual, then an object that might have been declared of type base class and instance of child class would simply call the base class destructor without calling the derived class destructor.
Hence, by making the destructor in the base class virtual, we ensure that the derived class destructor gets called before the base class destructor.
class a
{
public:
a(){printf("nBase Constructorn");}
~a(){printf("nBase Destructorn");}
};
class b : public a
{
public:
b(){printf("nDerived Constructorn");}
~b(){printf("nDerived Destructorn");}
};
int main()
{
a* obj=new b;
delete obj;
return 0;
}
Output:
Base Constructor
Derived Constructor
Base Destructor
By Changing
~a(){printf("nBase Destructorn");}
to
virtual ~a(){printf("nBase Destructorn");}
Output:
Base Constructor
Derived Constructor
Derived Destructor
Base Destructor
Hence, by making the destructor in the base class virtual, we ensure that the derived class destructor gets called before the base class destructor.
class a
{
public:
a(){printf("nBase Constructorn");}
~a(){printf("nBase Destructorn");}
};
class b : public a
{
public:
b(){printf("nDerived Constructorn");}
~b(){printf("nDerived Destructorn");}
};
int main()
{
a* obj=new b;
delete obj;
return 0;
}
Output:
Base Constructor
Derived Constructor
Base Destructor
By Changing
~a(){printf("nBase Destructorn");}
to
virtual ~a(){printf("nBase Destructorn");}
Output:
Base Constructor
Derived Constructor
Derived Destructor
Base Destructor
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