Virtual Destructor
Last Updated :
20 Feb, 2023
Deleting a derived class object using a pointer of base class type that has a non-virtual destructor results in undefined behavior. To correct this situation, the base class should be defined with a virtual destructor.
For example, the following program results in undefined behavior.
CPP
#include <iostream>
using namespace std;
class base {
public :
base()
{ cout << "Constructing base\n" ; }
~base()
{ cout<< "Destructing base\n" ; }
};
class derived: public base {
public :
derived()
{ cout << "Constructing derived\n" ; }
~derived()
{ cout << "Destructing derived\n" ; }
};
int main()
{
derived *d = new derived();
base *b = d;
delete b;
getchar ();
return 0;
}
|
Output
Constructing base
Constructing derived
Destructing base
Making base class destructor virtual guarantees that the object of derived class is destructed properly, i.e., both base class and derived class destructors are called. For example,
CPP
#include <iostream>
using namespace std;
class base {
public :
base()
{ cout << "Constructing base\n" ; }
virtual ~base()
{ cout << "Destructing base\n" ; }
};
class derived : public base {
public :
derived()
{ cout << "Constructing derived\n" ; }
~derived()
{ cout << "Destructing derived\n" ; }
};
int main()
{
derived *d = new derived();
base *b = d;
delete b;
getchar ();
return 0;
}
|
Output
Constructing base
Constructing derived
Destructing derived
Destructing base
As a guideline, any time you have a virtual function in a class, you should immediately add a virtual destructor (even if it does nothing). This way, you ensure against any surprises later.
Reference: Secure Coding
Share your thoughts in the comments
Please Login to comment...