Notify not getting the thread out of wait state

Question

I am trying to use 2 threads. 1 thread prints only odd number and the other thread prints only even number and It has to be an alternative operation.

Eg:

Thread1 1 Thread2 2 Thread1 3 Thread2 4 and so on..

Below is the program, please let me know where I am going wrong as the thread1 is not coming out of wait state even when the thread2 is notifying it..

 public class ThreadInteraction { public static void main(String[] args) { new ThreadInteraction().test(); } private void test() { ThreadA ta = new ThreadA(); Thread t = new Thread(ta); t.start(); try { Thread.sleep(1000); } catch (InterruptedException e1) { e1.printStackTrace(); } for(int i=2;i<=50;){ System.out.println("Thread2 "+i); synchronized (t) { try { t.notify(); t.wait(); } catch (Exception e) { e.printStackTrace(); } } i=i+2; } } } class ThreadA implements Runnable{ @Override public void run() { for(int i=1;i<50;){ System.out.println("Thread1 "+i); synchronized (this) { try { notify(); wait(); } catch (InterruptedException e) { e.printStackTrace(); } } i=i+2; } } }

Lokesh · Accepted Answer · 2013-10-06 07:26:42Z

Problem is that in one case you are taking lock on Thread t [synchronized (t) ] while in other case you are taking lock on TheadA object itself [synchronized(this)].

If you want threads to talk to each other then both should take lock on same object only then wait notify will work as you expect.

Edit:

There is another problem in your program, you are not using any variable to coordinate between 2 threads. SO you may see output like this 2,1,4,3...so on. Point is threads will work alternately but not in sequence. So you should share a single variable between 2 threads which should be incremented. Second issue is you are not taking care of spurious wake up calls [read some docs on this], you should always have wait called inside a while loop.

Thanks Lokesh. How dumb of me. I dont know how I couldn't recognize it. Yes it works now. Instead of getting a lock on t, I am taking a lock on ThreadA now. Thanks again

user001 · Accepted Answer · 2013-10-06 07:24:53Z

Modified my code based on the answer provided by Lokesh

public class ThreadInteraction { public static void main(String[] args) { new ThreadInteraction().test(); } private void test() { ThreadA ta = new ThreadA(); Thread t = new Thread(ta); t.start(); try { Thread.sleep(1000); } catch (InterruptedException e1) { e1.printStackTrace(); } for(int i=2;i<=50;){ System.out.println("Thread2 "+i); synchronized (ta) { try { ta.notify(); ta.wait(); } catch (Exception e) { e.printStackTrace(); } } i=i+2; } } } class ThreadA implements Runnable{ @Override public void run() { for(int i=1;i<50;){ System.out.println("Thread1 "+i); synchronized (this) { try { notify(); wait(); } catch (InterruptedException e) { e.printStackTrace(); } } i=i+2; } } }

Peter Lawrey · Accepted Answer · 2013-10-06 08:07:07Z

You have a real confusion of threads and locks. I suggest you create one and only one object to use for locking to start with as you don't appear to have a clear idea what you are locking.

If you notify() and nothing is listening, the signal is lost. However, a wait() can wake spuriously.

For this reason, a notify() should be accompanied by a state change and a wait() should be in a loop checking that change.

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