Question

Is the Wikipedia example at this article about method overiding in C++ correct?

Please see the code below where I have mentioned a comment //INCORRECT

Slightly confused about overiding in C++ and run time polymorphism. What should this cast do?

#include <iostream>

class Rectangle {
public:
    explicit Rectangle(double l, double w) : length(l), width(w) {}
    virtual void print() const;

private:
    double length;
    double width;
};

void Rectangle::print() const { // print() method of base class
   std::cout << "Length = " << this->length << "; Width = " << this->width;
}

class Box : public Rectangle {
public:
    explicit Box(double l, double w, double h) : Rectangle(l, w), height(h) {}
    virtual void print() const; // virtual is optional here, but it is a good practice to remind it to the developer

private:
    double height;
};

void Box::print() const { // print() method of derived class
   Rectangle::print();  // Invoke parent print() method.
   std::cout << "; Height= " << this->height;
}

int main(int argc, char** argv) {
   Rectangle rectangle(5.0, 3.0);   rectangle.print();
   // outputs:
   // Length = 5.0; Width = 3.0

   Box box(6.0, 5.0, 4.0);
   // the pointer to the most overridden method in the vtable in on Box::print
   //box.print(); // but this call does not illustrate overriding

   static_cast<Rectangle&>(box).print(); // this one does   

   // outputs:
   // Length = 5.0; Width = 3.0; Height= 4 // INCORRECT  

   //But it actually outputs Length = 6; Width = 5; Height = 4

   getchar();
   return 0;

}

Answer 1

this line: static_cast<Rectangle&>(box).print(); has same effect as this code would have:

Rectangle & r = box;
r.print();

and also same effect as this code would have:

Rectangle * r = &box;
r->print();

It means that Box object was created (Box box(6.0, 5.0, 4.0); = in runtime you see Box) and it doesn’t matter what is the type of pointer / reference where you store this object. The fact that you store box as a Rectangle* is what you see in compile time.

Most important fact about overriding virtual methods is that about “which method will be called?” is decided in runtime.

Hope this helps.