Java overloading method selection

Question

I'm trying to get my head round how Java selects which method is executed:

//Example 1 prints Square:add(Figure)
Figure fs = new Square();
fs.add(fs);

//Example 2 prints Square:add(Figure)
Rectangle rs = new Square();
rs.add(fs);

//Example 3 prints Rectangle:add(Rectangle). Expected Square:add(Square)
rs.add(new Square());

//Example 4 prints Rectangle:add(Rectangle). Expected Square:add(Figure)
Square ss = new Square();
ss.add(rs);

class Figure
{
    public void add(Figure f){ System.out.println("Figure:add(Figure)"); }
}

class Rectangle extends Figure
{
    @Override
    public void add(Figure f){ System.out.println("Rectangle:add(Figure)"); }
    public void add(Rectangle r){ System.out.println("Rectangle:add(Rectangle)"); }
}

class Square extends Rectangle
{
    @Override
    public void add(Figure f){ System.out.println("Square:add(Figure)"); }
    public void add(Square s){ System.out.println("Square:add(Square)"); }
}

What I've learned here is

Method signature gets determined based on compile time data types
Actual method invoked depends on the dynamic type of the object the method is called on.

Based on that, the result of the first two calls is as expected. However, I don't understand the result of example 3 and 4.

It seems to be specified in the java language specification, but I don't understand it.

As an author of this assignment question I can confirm that the provided answers are correct. — Dominik, Mar 12 '17 at 20:56

score 20 · Accepted Answer · answered Feb 03 '13 at 19:37

However, I don't understand the result of example 3 and 4.

Okay, let's look at them individually.

Example 3

//Example 3 prints Rectangle:add(Rectangle). Expected Square:add(Square)
rs.add(new Square());

The important parts are the compile-time types of the expressions rs and new Square().

rs is only declared as Rectangle, so the compiler will look at the methods declared by Rectangle and its superclasses:

public void add(Figure f)
public void add(Rectangle r)

The type of the expression new Square() is Square, so both methods are applicable - but the second one is more specific.

So the compiler will call add(Rectangle) on the object that rs refers to. That's it for the compile-time side.

At execution time, the value of rs refers to an instance of Square - but Square doesn't override add(Rectangle) so the method picked is the implementation in Rectangle:

public void add(Rectangle r){ System.out.println("Rectangle:add(Rectangle)"); }

Example 4

//Example 4 prints Rectangle:add(Rectangle). Expected Square:add(Figure)
Square ss = new Square();
ss.add(rs);

Again, let's consider the compile-time types involved... ss is of type Square, and rs is of type Rectangle (compile-time types, remember).

The methods declared by Square and its superclasses are:

public void add(Figure f)
public void add(Rectangle r)
public void add(Square s)

As the compile-time type of rs is only Rectangle (not Square), the first two methods are applicable, but the third isn't. Therefore, again, add(Rectangle) is picked at compile time (as it's more specific than add(Figure)).

Again, the execution time type of ss is Square, which doesn't override add(Rectangle), so the implementation in Rectangle is used.

Let me know if anything in here is confusing - if you can be specific about which part, that would be great.

score 4 · Answer 2 · answered Feb 03 '13 at 19:40

rs.add(new Square());

The declared type of rs is Rectangle. So it looks at a method in Rectangle and all superclasses taking a Square or a type compatible with Square as argument. The most specific one is add(Rectangle) since Square is a Rectangle, and since Rectangle is more specific than Figure.

Square ss = new Square();
ss.add(rs);

Looks for a method add(Rectangle) in Square and all super-classes. Rectangle.add(Rectangle) is chosen, since Square.add(Square) doesn't apply (a Rectangle is not a Square), and Square.add(Figure) is less specific.

Java overloading method selection

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