How to make thread communication using deque?

Question

I writing a program, in which One thread is for inserting elements in to Deque, another is for reading elements from same Deque. For that I am doing like this,

public class ThreadCommunicationInDeque {
public static void main(String args[])
{
    Deque deque=new LinkedList<>();
    InsertingThread it=new InsertingThread(deque);
    ReadingThread1 rt=new ReadingThread1(deque);
    it.start();
    rt.start();
}
}

class InsertingThread extends Thread{

Deque deque=new LinkedList<>();
InsertingThread(Deque deque){
    this.deque=deque;
}

@Override
public void run(){
   for(int i=0;i<50;i++){
       try {
           deque.add(i);
           System.out.println(deque.getLast());
           Thread.sleep(100);
       } catch (InterruptedException ex) {
           Logger.getLogger(InsertingThread.class.getName()).log(Level.SEVERE, null, ex);
       }
   }
}

}
class ReadingThread1 extends Thread{

Deque deque=new LinkedList<>();

ReadingThread1(Deque deque){
  this.deque=deque;
}

@Override
public void run(){
    if(deque.isEmpty()){
        System.out.println("No elements in queue");
    }else
    {
        System.out.println(deque.getLast());
        try {
            Thread.sleep(100);
        } catch (InterruptedException ex) {
            Logger.getLogger(ReadingThread1.class.getName()).log(Level.SEVERE, null, ex);
        }
    }

}
}

My problem is Here InsertingThread is working fine but ReadingThread is stops by displaying No elements in queue. I am not getting where I am doing wrong , Can anyone tell me how to fix it. And tell me is there any better way to do this?

Answer 1

That's because you don't have consumer running in a loop like a thread should be. It just runs for once and see's that there are no elements in queue (which might be true if say consumer was the thread that ran first) and then exits safely from the thread.

To begin with try something in your consumer's run method like and then fine tune it if you need:

while (true) {
        if (deque.isEmpty()) {
            System.out.println("No elements in queue");
        } else {
            System.out.println("consumed " + deque.removeLast());//note get will still have queue populated so remove seems better?
            try {
                Thread.sleep(100);
            } catch (InterruptedException ex) {
            }
        }
}

By the way, you are unnecessary creating instance of dequeue in your producer and consumer. You could just remove the instantiation i.e. from

Deque<Integer> deque = new LinkedList<Integer>();

To

Deque<Integer> deque;

if you just want to display last element from within a thread (You could just do it from main method though), then you could do it by calling join method as below (so it lets producer complete and then start consumer thread) in your main method:

public static void main(String args[]) throws InterruptedException {
    Deque<Integer> deque = new LinkedList<Integer>();
    InsertingThread it = new InsertingThread(deque);
    ReadingThread1 rt = new ReadingThread1(deque);
    it.start();
    it.join();
    rt.start();
}

Answer 2

Order in which threads are executed is random and cannot be determined. So starting producer thread it.start() does not mean insertions will always be before you retrieval rt. start(). Since you will be inserting. For retrieving you should try replacing code with one given by almas.