Java method selection with overloading and overriding

Question

I am having hard time to understand some principles about overriding and overloading.

public class Figure{
    public void stampa(Figure f){
        System.out.println("Figure");
    }
}
public class Square extends Figure{
    public void stampa(Square q){
        System.out.println("Square");
    }
}
public class Rectangle extends Square{
    public void stampa(Rectangle r){
        System.out.println("Rectangle");
    }
    public void stampa(Square r){
        System.out.println("Particular Rectangle");
    }
    public void stampa(Figure f){
        System.out.println("Particular Figure");
    }
}

public class Test{
    public static void main(String args[]){

        Figure f1,f2;
        Square q = new Rectangle();
        Rectangle r = new Rectangle();
        f1 = new Square();
        f2 = new Rectangle();

        f1.stampa(f2); //Figure
        q.stampa(r); //Particular Rectangle 
        f1.stampa(q); //Figure
        q.stampa(f1); //Particular Figure 
        q.stampa(q); //Particupar Rectangle
    }
}

I know that public void stampa(Square q) is overloading public void stampa(Figure f) and not overrding it.

And public void stampa(Rectangle r) and public void stampa(Figure f) are also overloading public void stampa(Square q).

Also public void stampa(Square q) in Rectangle class is overriding the method in Square class.

First question

it's about this result : q.stampa(f1); //Particular Figure I know that at compile-time q is an Square so i will look at this method public void stampa(Square q)in Square class. and at run-time q is Rectangle so i thought the result might be "Particular Rectangle" instead of "Particular Figure"

Not sure what I'm doing wrong

Second question

If at this moment Rectangle extends Figure and not any more Square, I will certainly have compilation error on Square q = new Rectangle(); what happens to the varibale q (there will be have such a variable as Square q or we dont have any varible with name q?) and what will be the result of q.stampa(f1);

Thanks and sorry for my english and please correct me if I'm wrong at some point.

Your class is defined as `Figura`. Unless this is just a copy and paste typo, this would create some problems. — kunruh, Jan 13 '17 at 21:11
Besides your override/overload question, I would like to comment that the inheritance hierarchy is wrong, as a Rectangle is not a special Square, otherwise every rectangle would also be (is-a) a square, and this is not true. — Dominik, Mar 12 '17 at 21:03

score 4 · Answer 1 · answered Jan 13 '17 at 21:08

First question:

q's declared type is Square. f1's declared type is Figure. So the compiler looks for a method named stampa, in the class Square and its superclasses, if a method accepting a Figure (or a superclass of Figure) exists. It finds onein Figure: public void stampa(Figure f). At compile time, what matters is the declared type of the variable.

Now, at runtime, the actual runtime type of the object referred by q is examined: it's a Rectangle. So the runtime looks for a method public void stampa(Figure f) that overrides the Figure method, in Rectangle, and all its superclasses until Figure. That's what allows polymorphism. It finds one in Rectangle:

public void stampa(Figure f) {
    System.out.println("Particular Figure");
}

Second question:

You will indeed have a compilation error, so reasoning about the type of the variable q doesn't make sense: there will be no such variable, since the code does not compile, and the class is thus not generated by the compiler.

shmosel · Accepted Answer · 2017-01-13T21:28:56.410

I know that at compile-time q is an Square so i will look at this method public void stampa(Square q)in Square class.

You're forgetting that Square contains two methods: stampa(Square q) and stampa(Figure f). The latter is inherited from Figure and is selected at compile time as the ~~most~~ only appropriate overload.

and at run-time q is Rectangle so i thought the result might be "Particular Rectangle" instead of "Particular Figure"

Overloads are selected at compile time, and overrides at runtime. As stated, stampa(Figure f) is already selected at compile time. Once it's actually invoked, the runtime sees that q is an instance of Rectangle and delegates the invocation to Rectangle.stampa(Figure f).

As for the second question, I don't really understand what you're asking. A compile error means the code is invalid and nothing happens. q.stampa(f1) will never be called and will never return a result.

Java method selection with overloading and overriding

2 Answers2