Timed while loop not terminating

Question

Running this code, I would expect it to increment the test variable for 5 seconds and then finish.

import java.util.Timer;
import java.util.TimerTask;

public class Test {
    private static boolean running;

    public static void main( String[] args ) {
        long time = 5 * 1000;       // converts time to milliseconds
        long test = Long.MIN_VALUE;
        running = true;

        // Uses an anonymous class to set the running variable to false
        Timer timer = new Timer(); 
        timer.schedule( new TimerTask() { 
            @Override
            public void run() { running = false; }
        }, time );

        while( running ) {
            test++;
        }

        timer.cancel();
        System.out.println( test );
    }
}

However when I run it the program doesn't end (I assume, I have given it a reasonable amount of time). However if I change the while loop to

    while( running ) {
        System.out.println();
        test++;
    }

The program finishes in the expected amount of time (and prints out a lot of lines). I don't understand. Why does this behaviour occur?

I just tested your code using JDK 7 with IntelliJ 11 and both versions finished in a few seconds. What are you using to run? — Tim Biegeleisen, May 27 '15 at 05:28
@AndyTurner wow, I've never come across the volatile keyword before. Thats makes the code work. Thanks! — DenverCoder9, May 27 '15 at 05:36

Tagir Valeev · Accepted Answer · 2015-05-27T05:37:40.553

According to Java Memory Model there's no guarantee when non-volatile field will be visible from another thread. In your case your main method is JIT compiled and JIT-compiler reasonably assume that as current thread does not modify the running variable in a loop, then we should not bother reading it on every iteration and convert the loop to the infinite one. There's even FindBugs warning regarding this case.

When you add a System.out.println call, seems that JIT compiler cannot inline it, so it cannot be sure that this method does not modify the running variable, thus it switches off the field reading optimization. However that should not be considered as problem solution: it's possible that newer versions of Java will be smarter and optimize your loop even if you have System.out.println inside.

I accepted this answer because it links additional information. Thanks! — DenverCoder9, May 27 '15 at 05:38

score 2 · Answer 2 · answered May 27 '15 at 05:33

Lets get closer into your code...

If you debug your code, it will work. That's because you will only see one thread and no cache. Java can use thread-based cache mechanisms. You need to read and write it into main memory.

So if you use volatile keyword on your running variable, jvm will see it as editable by multiple threads, and it shouldn't be cached.

Therefore in your situation solution is adding volatile to your running variable.

score 2 · Answer 3 · answered May 27 '15 at 05:35

You are updating the running variable in a different thread from your main method. The Java Memory Model is such that you are not guaranteed to see updates to non-volatile variables in different threads.

The easiest solution is to make the variable volatile: this forces the 'current' value of the variable to be checked when the thread accesses it.

Other solutions include using AtomicBoolean instead of boolean, and wrapping access to running in mutually synchronized blocks (i.e. the code that accesses running is synchronized on the same monitor as the code which updates it).

I strongly recommend that you pick up a copy of Java Concurrency In Practice, which describes this problem in detail.

Timed while loop not terminating

3 Answers3