illustrating volatile : is this code thread-safe?

Question

I\'m trying to illustrate the use and importance of volatile with an example that would really not give a good result if volatile was omitted.

Answer 1

Victor is right, there are issues with your code: atomicity and visibility.

Here's my edition:

    private int count;
    private volatile boolean stop;
    private volatile boolean stopped;

    @Override
    public void run() {
        while (!stop) {
            count++; // the work
        }
        stopped = true;
        System.out.println("Count 1 = " + count);
    }

    public void stopCounting() {
        stop = true;
        while(!stopped)
           ; //busy wait; ok in this example
    }

    public int getCount() {
        if (!stopped) {
            throw new IllegalStateException("not stopped yet.");
        }
        return count;
    }

}

If a thread observes that stopped==true, it's guaranteed that the work completes and the result is visible.

There is a happens-before relation from volatile write to volatile read (on the same variable), so if there are two threads

   thread 1              thread 2

   action A
       |
 volatile write  
                  \
                     volatile read
                          |  
                       action B

action A happens-before action B; writes in A are visible by B.