How Java synchronize updates thread value?

Question

I have next code:

boolean signal;

@Test
public void test() throws InterruptedException {
    Thread thread = new Thread(new Runnable() {
        @Override
        public void run() {
            while (!signal){
                // empty loop body
            }
        }
    });
    thread.start();
    Thread.sleep(1000);
    signal = true;
    thread.join();
}

It runs infinity loop due to creation of local copy of signal variable in thread. I know that I can fix it by making my signal variable volatile. But also loop can successfully exit if add synchronized block inside my loop (even empty):

boolean signal;

@Test
public void test() throws InterruptedException {
    Thread thread = new Thread(new Runnable() {
        @Override
        public void run() {
            while (!signal){
                synchronized (this) {

                }
            }
        }
    });
    thread.start();
    Thread.sleep(1000);
    signal = true;
    thread.join();
}

How synchronized updates my signal value inside thread?

It would be perfectly legal for your second program to never exit - your JVM/CPU architecture seems to apply a stronger memory model than what the JLS requires. — assylias
– assylias, Commented Jul 8, 2016 at 15:02
Interesting read: shipilev.net/blog/2016/close-encounters-of-jmm-kind — JB Nizet
– JB Nizet, Commented Jul 8, 2016 at 15:21
He shows a closely related example in shipilev.net/blog/2016/close-encounters-of-jmm-kind/… — assylias
– assylias, Commented Jul 8, 2016 at 15:32

Marco Vargas · Accepted Answer · 2016-07-08 15:13:51Z

1

Synchronized does not updates the signal value itself, it basically just places a couple of flags to avoid two threads use the same object at the same time; something like: MonitorEnter and MonitorExit.

The first one locks the object, and the second one releases.

Take a look at the following article: how-the-java-virtual-machine-performs-thread-synchronization.

Please notice the article is very old; but as far as I understand the logic behind remains.

answered Jul 8, 2016 at 15:13

Marco Vargas

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2 Comments

Cortwave Over a year ago

I know that, but why adding of synchronized block to loop updates signal variable in my second example?

Marco Vargas Over a year ago

Because the one without synchronized is a selfish thread; it means; it will stop until finishes its work; any thread can become a selfish thread if inside the loop it does not have a yield/sleep/Blocking Operation such I/O Now, when you placed the synchronized you make your selfish thread to be not selfish, since there is a blocking operation on the object "signal" The behavior itself depends a lot on the Infrastructure where you run it; but basically you have two threads there 1) main 2) thread and since the second is selfish it won't let main to continue...

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