Simplest and understandable example of volatile keyword in Java

Question

I\'m reading about volatile keyword in Java and completely understand the theory part of it.

But, what I\'m searching for is, a good case example, which sho

Answer 1

Please find the solution below,

The value of this variable will never be cached thread-locally: all reads and writes will go straight to "main memory". The volatile force the thread to update the original variable for each time.

public class VolatileDemo {

    private static volatile int MY_INT = 0;

    public static void main(String[] args) {

        ChangeMaker changeMaker = new ChangeMaker();
        changeMaker.start();

        ChangeListener changeListener = new ChangeListener();
        changeListener.start();

    }

    static class ChangeMaker extends Thread {

        @Override
        public void run() {
            while (MY_INT < 5){
                System.out.println("Incrementing MY_INT "+ ++MY_INT);
                try{
                    Thread.sleep(1000);
                }catch(InterruptedException exception) {
                    exception.printStackTrace();
                }
            }
        }
    }

    static class ChangeListener extends Thread {

        int local_value = MY_INT;

        @Override
        public void run() {
            while ( MY_INT < 5){
                if( local_value!= MY_INT){
                    System.out.println("Got Change for MY_INT "+ MY_INT);
                    local_value = MY_INT;
                }
            }
        }
    }

}

Please refer this link http://java.dzone.com/articles/java-volatile-keyword-0 to get more clarity in it.