Display difference between volatile and usual variable in Java

Question

I am trying to create an example to display the difference between volatile and usual variables like:

package main;

public class TestVolatile extends Thread {

    public int l = 5;
    public volatile int m = -1;

    public TestVolatile(String str) {
        super(str);
    }

    public void run() {
        int i = 0;
        while ((l > 1) && (l < 10)) {
            if (m >= 0) {
                m++;
            }
            i++;
            l = 5;
            System.out.println("5=" + i + " m=" + m);
        }

    }

    public static void main(String[] args) throws InterruptedException {
        TestVolatile tva = new TestVolatile("ThreadA");

        tva.start();
        sleep(5);
        synchronized (tva) {
            tva.m = 5;
            tva.l = 10;
        }
    }

}

So m is volatile, l is not. I suppose that exiting from the while loop depends on the value of l.
Because the value of l is not volatile - m will be incremented at least 1 time after l has been assigned 5. But I have run the code 10 times and always m==5.
So I suppose that I am wrong. How to fix this problem? Thank you.

Thanks for answers, but not all run well. I set like:

volatile int x = 0;
volatile int y = 0;

So now the variables have to be the same! But that is not the case.

x: 346946234 y: 346946250
x: 346946418 y: 346946422
x: 346946579 y: 346946582
x: 346946742 y: 346946745
x: 346946911 y: 346946912

I've added a program to my answer, which will probably show the difference. — jmg, Apr 20 '11 at 10:27
Maybe relevant: http://stackoverflow.com/questions/3964317/memory-barriers-and-coding-style-over-a-java-vm/3964347 — andersoj, Apr 20 '11 at 13:15

jmg · Accepted Answer · 2011-04-20T14:20:55.327

You are synchronizing the main thread and your test thread. Therefore Java guarantees to make any changes visible performed by the other thread.

Btw, it is impossible to construct an example which deterministically shows a difference between volatile and non-volatile. The best you can hope is to get a program which shows the difference with a quite high probability. If the threads run interleaved on the same core. You won't be able to show any difference at all.

The following program shows on my computer the difference between volatile and non-volatile variables.

public class ShowVolatile {

    final static int NUM_THREADS = 1;

    int x = 0;
    volatile int y = 0;

    public static void main(String... args) {

        final ShowVolatile sv = new ShowVolatile();

        for (int i=0; i< NUM_THREADS; i++) {
            new Thread(new Runnable() {
                public void run() {
                    while (true) {
                        sv.x += 1;    
                        sv.y += 1;    
                    }
                }
            }).start();
        }


        while (true) {
            System.out.println("x: " + sv.x + " y: " + sv.y);
        }
    }

}

If you increase the number of threads you will see additional synchronization misses. But a thread count of 1 is enough. At least on my hardware a Quad-Core i7.

Thank you. When set NUM_THREADS e.g. 15 difference increase. Always volatile variable more than usual because usual always lost. — user710818, Apr 20 '11 at 12:42
But System.out.println("x: " + sv.x + " y: " + sv.y); suppose not correct because is possible that main thread read sv.x, after thread incremented sv.y and after read sv.y and print. — user710818, Apr 20 '11 at 13:06
@user710818: Yes, but that would only explain a difference of 1. Would you consider accepting this answer? — jmg, Apr 20 '11 at 14:22

score 2 · Answer 2 · edited Jul 18 '12 at 13:05

2

Having multiple threads doing concurrent non-synchronized ++ on your volatile variable should help. However it won't ever be deterministic. It might also depend on the underlying OS and CPU.

edited Jul 18 '12 at 13:05

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answered Apr 20 '11 at 09:50

Guillaume

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Display difference between volatile and usual variable in Java

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