Thread join() does not wait

Question

I'm trying to learn about threads and I do not understand the join() method.

I have a Thread (ThreadAdd.java) which adds 1 to a static int.

public class ThreadAdd extends Thread{

public static int count;

    @Override
    public void run() {

        try {
            Thread.sleep(100);
        } catch (InterruptedException ex) {
            Logger.getLogger(ThreadAdd.class.getName()).log(Level.SEVERE, null, ex);
        }
        ThreadAdd.count++;

    }

}

In my main method I launch 2 threads :

 public static void main(String[] args) throws InterruptedException {

   ThreadAdd s1 = new ThreadAdd();
   ThreadAdd s2 = new ThreadAdd();
   s1.start();s2.start();
   s1.join();
   s2.join();
   System.out.println(ThreadAdd.count);

}

I do not understand why most of the time the result is 2 but sometimes it returns 1.

Answer 1

The reason why you sometimes see 1 is not because join() fails to wait for the thread to finish, but because both threads tried to modify the value concurrently. When this happens, you may see unexpected results: for example, when both threads try to increment count which is zero, they both could read zero, then add 1 to it, and store the result. Both of them will store the same exact result, i.e. 1, so that's what you are going to see no matter how long you wait.

To fix this problem, add synchronized around the increment, or use AtomicInteger:

public static AtomicInteger count = new AtomicInteger(0);
@Override
public void run() {
    try {
        Thread.sleep(100);
    } catch (InterruptedException ex) {
        Logger.getLogger(ThreadAdd.class.getName()).log(Level.SEVERE, null, ex);
    }
    ThreadAdd.count.incrementAndGet();
}

Answer 2

The join method is not the real issue here. The problem is that your counter is not prepared for interthread synchronization, which may lead to each thread observing a different value in count.

It is highly recommended that you study some topics of concurrent programming, including how it is handled in Java.