Thread: function = 0 ;

Basically, it is saying that it will never be used
That's completely wrong.

It's saying that it requires an interface with that function name, with those parameters, but it requires derived objects to create their own implementation of that function. Eg. a class called Shape should have a function that would output it's area... but it doesnt know what type of shape it is so it declares a pure virtual function called Area() that the derived objects must implement themselves.. (since each shape will have it's own way of calculating the area).

This doesn't mean that the Shape object can't call the Area() function.

A class such as this, that contains a pure virtual function (ie function = 0), can not be instantiated.

U.

Sorry about my bad explanation, I meant that it will never be used by itself and that it can only be used from a derived class
this is still wrong mate. Just because a class has a pure virtual function, it doesnt mean that it can't call that function itself!

Here is an example of a class that calls one of it's own pure virtual functions.. and it works fine (BTW, i havent' compiled this... but i know it works)

Code:

class Shape
{
public:
   void ShowArea(void);
   virtual int Area(void) = 0;
};

class Square : public Shape
{
public:
   virtual int Area(void);
};

void Shape::ShowArea(void)
{
   cout << "My area is " << Area() << endl;
}

int Square::Area(void)
{
    return(side * side);
}

i hope this clears it up.
U.