CODEWITHSUNDEEP
javavolatile
Bala
Bala

Reputation: 39

Java Volatile Variable

I am trying to understand volatile usage by the below example. I expect it to print 10 first and then 15 second. But most of the time i end getting 10 and 10. Is some thing with the below code itself.

class T implements Runnable {

    private volatile int x = 10;

    @Override
    public void run() {
        if(x==10) {
            System.out.println(x);
            x = x+ 5;
        } else if(x==15) {
            System.out.println(x);
        }
    }
}


public class Prep {

    public static void main(String [] args) {
        T t1 = new T();
        new Thread(t1).start();
        new Thread(t1).start();
    }
}

Upvotes: 2

Views: 129

Answers (2)

JB Nizet
JB Nizet

Reputation: 692071

You just have a race condition: both threads run in parallel, and thus you have (for example)

  • thread 1 tests if x == 10 (true)
  • thread 2 tests if x == 10 (true)
  • thread 1 prints 10
  • thread 2 prints 10
  • ...

The fact that x is volatile here is irrelevant. The only thing that volatile guarantees in this example is that if thread 1 has already incremented x when thread 2 reads its value, then thread 2 will see the incremented value. But it doesn't mean that the two threads can't run in parallel as shown above.

Upvotes: 7

TheLostMind
TheLostMind

Reputation: 36304

This will work :

public static void main(String[] args) {
        T t1 = new T();
        new Thread(t1).start();
        try {
            Thread.currentThread().sleep(1000); // sleeping for sometime .
        } catch (InterruptedException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        new Thread(t1).start();
    }

answer : 10 15

Reason : The operations 

if(x==10) {
            System.out.println(x);

might have executed first for the first thread, then context would have switched back to Thread2 (race condition as JB Nizet explains..) so when both threads print the value of x, it is still 10.

Upvotes: 0

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