问题

I am trying to achieve this: Created two different threads, one prints odd numbers, one prints even numbers. Once one thread prints a number, it has to wait for the other thread and so on, that is one-after-other.

To achieve this, i am using synchronized block along with wait() and notify().

I am creating a class whose's object will be used to pass to synchronized block in both the threads.

Here is the code:

--> This is used object which will be passed to synchronized block.

package com.vipin.multithread.variousdemos;

    public class SyncObject {

        public SyncObject () {  

        }
}

Odd Thread:

package com.vipin.multithread.variousdemos;

public class OddThread implements Runnable {

private Thread t;
int index=0;
SyncObject so=null;

int odd_nums[] = {1,3,5,7,9};

public OddThread(SyncObject so) {
    t = new Thread(this,"Odd Thread");
    this.so = so;
    t.start();
}

public Thread getThreadInstance() {
    return t;
}

@Override
public void run() {
    while (true) {
        synchronized(so) {
            System.out.println("Odd num is --->" + odd_nums[index]);
            try {
                so.wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            index++;
            so.notify();
            if(index>=5) {
                return;
            }
        }
    }
}
}

Even Thread: UPDATE

package com.vipin.multithread.variousdemos;

public class EvenThread implements Runnable {

private Thread t;
int index=0;
SyncObject so=null;

int even_nums[] = {2,4,6,8,10};

public EvenThread(SyncObject so) {
    t = new Thread(this, "Even thread");
    this.so = so;
    t.start();
}

public Thread getThreadInstance() {
    return t;
}

@Override
public void run() {
    while(true) {   
        synchronized(so) {
            System.out.println("Even num is --->" + even_nums[index]);
            so.notify(); <-- Here we are notifying.
            try {
                so.wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            index++;
            //so.notify(); <-- commented out.
            if(index>=5) {
                break;
            }
        }
    }
}
}

Main Application:

package com.vipin.multithread.variousdemos;

public class EvenOddDemo {

public static void main(String[] args) throws InterruptedException {
    SyncObject so = new SyncObject();

    OddThread ot = new OddThread(so);
    EvenThread et = new EvenThread(so);

    System.out.println("\nIn main thread");

    Thread.sleep(1000000000);

    System.out.println("Exiting main thread...");
}
}

---> As seen in the code, I am creating two threads to print even and odd numbers. I am using synchronized block, and passing object of type ==> SyncObject.

SyncObject I am passing as argument to these different threads in main.

However, this programs halts, i.e stuck only first statement gets executed, and then it waits forever:

Here is the output:

Odd num is --->1

In main thread Even num is --->2

I am not able to understand why this program waits for ever, I am using SAME object on which we are invoking synchronized(), wait() and notify(). As per my understanding, it should work, not sure why this is not working.

Any clues as to why this is waiting forever.

UPDATE:

I did some changes in the code, UPDATE and it works fine.

I still have some doubt. Does notify() be called by the thread even if it has not locked the monitor, like in my case after i updated the code.

Sequence of events:

Odd thread gets executed first, then it calls wait() <-- it releases the monitor and now in sleep mode.

Even thread runs, prints msg, and calls notify() <-- here i am not having clear understanding.

When Even thread calls notify(), at that point it has the monitor, so when it calls notify(), does is still own the monitor?

Now, after Even thread calls notify(), then Odd thread gets notified, and hence it starts execution from the point it was sleeping. It is doing some execution and calls notify(), at that points I presume Odd thread is NOT owning the monitor, it calls notify(). So, my question is, does notify() work same whether or not the thread owns the monitor?

It is only when one do the code, one really understands this. I read book and i felt i understood everything, and seems i am back to square one!

回答1:

The problem here is simply that both threads go straight into wait. Thread 1 gets so, prints value then waits. Thread 2 then gets so, prints value then waits. So both are sleeping away, since nobody is there to notify them. So, a simple fix would be to do so.notify(), right before so.wait(). Then they're not infinitely waiting.

EDIT

Odd thread starts, executes & then waits. Then even thread starts, executes, notifies & then waits. Even thread holds the lock over the monitor until it goes into wait.

When the even thread called on notify, the odd thread awakens & polls for the lock. Once the even thread goes into wait (& releases the lock), then the odd thread can obtain the lock.

If the even thread had not called on notify, then the odd thread would continue to sleep. The even thread would have gone to wait & released the lock. No thread is polling or attempting to obtain the lock, hence the program remains in the suspended state.

The documentation also provides a similar explanation. I hope that clears your doubts.

来源：https://stackoverflow.com/questions/30141414/program-gets-halted-wait-and-notify

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