What does 'has virtual method … but non-virtual destructor' warning mean during C++ compilation?

Question

#include 
using namespace std;

class CPolygon {
  protected:
    int width, height;
  public:
    virtual int area ()
      { return (0); }
  };

cl         


        

Answer 1

If a class has a virtual method, that means you want other classes to inherit from it. These classes could be destroyed through a base-class-reference or pointer, but this would only work if the base-class has a virtual destructor. If you have a class that is supposed to be usable polymorphically, it should also be deletable polymorphically.

This question is also answered in depth here. The following is a complete example program that demonstrates the effect:

#include 

class FooBase {
public:
    ~FooBase() { std::cout << "Destructor of FooBase" << std::endl; }
};

class Foo : public FooBase {
public:
    ~Foo() { std::cout << "Destructor of Foo" << std::endl; }
};

class BarBase {
public:
    virtual ~BarBase() { std::cout << "Destructor of BarBase" << std::endl; }
};

class Bar : public BarBase {
public:
    ~Bar() { std::cout << "Destructor of Bar" << std::endl; }
};

int main() {
    FooBase * foo = new Foo;
    delete foo; // deletes only FooBase-part of Foo-object;

    BarBase * bar = new Bar;
    delete bar; // deletes complete object
}

Output:

Destructor of FooBase
Destructor of Bar
Destructor of BarBase

Note that delete bar; causes both destructors, ~Bar and ~BarBase, to be called, while delete foo; only calls ~FooBase. The latter is even undefined behavior, so that effect is not guaranteed.