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POJava overloading method selection
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copied!<p>I'm trying to get my head round how Java selects which method is executed:</p> <pre><code>//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)"); } } </code></pre> <p>What I've learned <a href="https://stackoverflow.com/questions/1572322/overloaded-method-selection-based-on-the-parameters-real-type">here</a> is</p> <ul> <li>Method signature gets determined based on compile time data types</li> <li>Actual method invoked depends on the dynamic type of the object the method is called on.</li> </ul> <p>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.</p> <p>It seems to be specified in the <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.12" rel="nofollow noreferrer">java language specification</a>, but I don't understand it.</p>

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