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virtual destructor in c++

In the code below, why is the ~Derived() destructor called automatically?

#include<iostream>
using namespace std;
class Base
{
public:
    virtual ~Base()
    {
        cout << "Calling ~Base()" << endl;
    }
};

class Derived: public Base
{
private:
    int* m_pnArray;

public:
    Derived(int nLength)
    {
        m_pnArray = new int[nLength];
    }

    virtual ~Derived()
    {
        cout << "Calling ~Derived()" << endl;
        delete[] m_pnArray;
    }
};

int main()
{
    Derived *pDerived = new Derived(5);
    Base *pBase = pDerived;
    delete pBase;

    return 0;
}
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1 Answer

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by (71.8m points)

Because your base class destructor is virtual

virtual ~Base();

the call to delete on a pointer to a base class results in virtual call to destructor and as any virtual call is dispatched to matching function in derived class. It is not only good, but necessary: otherwise the behavior is undefined.

This is crucial for a derived classes which destructor is not an empty function. Non-virtual call would otherwise result in calling base class destructor, derived resources being leaked, etc.


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