How method preference work in java?

Question

I just want to understand how below code snippet work ?

class AnnaThread extends Thread {

    public static void main(String args[]){
        Thread t = new AnnaThread();
        t.start();

    }

    public void run(){
        System.out.println("Anna is here");
    }
    public void start(){
        System.out.println("Rocky is here");
    }
}

Output - Rocky is here

Answer 1

There's not much to explain.

• You override start() with code that prints Rocky is here
• then you call start() which prints Rocky is here.
• (the run method is never involved)

People often confuse the purpose of start and run. See for instance this question:

      Why we call Thread.start() method which in turns calls run method?

The rules are simple:

• Thread.run is an ordinary method (no magic)

• Thread.start contains some magic because it spawns a separate thread (and lets that thread invoke run).

  • If you override Thread.start with your own method, then there's no magic left anywhere.

Answer 2

what you have here is a Java class which extends the Thread class (http://docs.oracle.com/javase/7/docs/api/java/lang/Thread.html)

class AnnaThread extends Thread {

then in your main method you create a new instance of the class which is a Thread (since the class extends the Thread)

    public static void main(String args[]){
        Thread t = new AnnaThread();

then you call the method start which follows bellow

        t.start();

which prints

System.out.println("Rocky is here");

you could as well call the other method if you add the following line in your code

      t.run();

in which case the method run would be executed which would print

        System.out.println("Anna is here");