How do virtual destructors work?

Question

Few hours back I was fiddling with a Memory Leak issue and it turned out that I really got some basic stuff about virtual destructors wrong! Let me put explain my class design.

class Base
{
  virtual push_elements()
  {}
};

class Derived:public Base
{
vector<int> x;
public:
   void push_elements(){ 
      for(int i=0;i <5;i++)
         x.push_back(i); 
   }
};

void main()
{
    Base* b = new Derived();
    b->push_elements();
    delete b;
}

The bounds checker tool reported a memory leak in the derived class vector. And I figured out that the destructor is not virtual and the derived class destructor is not called. And it surprisingly got fixed when I made the destructor virtual. Isn't the vector deallocated automatically even if the derived class destructor is not called? Is that a quirk in BoundsChecker tool or is my understanding of virtual destructor wrong?

Answer 1

Deleting a derived-class object through a base-class pointer when the base class does not have a virtual destructor leads to undefined behavior.

What you've observed (that the derived-class portion of the object never gets destroyed and therefore its members never get deallocated) is probably the most common of many possible behaviors, and a good example of why it's important to make sure your destructors are virtual when you use polymorphism this way.