在多线程中使用等待和通知方法

Question

Now, I started to study Threading and decided to write this class to demonstrate the idea of Multi-Threading but the output wasn't what i want.

i wrote the class as follows : -

import java.util.LinkedList;
class QueueTest extends LinkedList{
    int capacity;
    LinkedList list;
    public QueueTest(int capacity){
        this.capacity = capacity;
        list = new LinkedList();
    }
    public synchronized void addElements(int i)throws InterruptedException{
        if(list.size() == capacity ){
                wait();
        }
        add(i);
        System.out.println("I added : "+i);
        notify();
    }
    public synchronized int getElements() throws InterruptedException{
        if(!isEmpty()){
                wait();
        }
        int i = (Integer) list.remove();
        System.out.println("I got : "+i);
        notify();
        return i;
    }
}
class Add implements Runnable{
    QueueTest t ;
    public Add(QueueTest t){
        this.t = t;
        new Thread(this,"Add").start();
    }
    @Override
    public void run(){
        int i = 0;
        while(t.size() <= t.capacity){
            try{
                t.addElements(i++);} catch(InterruptedException e){}
        }
    }
}
class Remove implements Runnable{
    QueueTest t ;
    public Remove(QueueTest t){
        this.t = t;
        new Thread(this,"Remove").start();
    }
    @Override
    public void run(){
        while(!t.isEmpty()){
            try{
            t.getElements();} catch(InterruptedException e){}
        }
    }
}
public class FullQueue{
    public static void main(String[] args){
        QueueTest t = new QueueTest(5);
        new Add(t);
        new Remove(t);
    }
}

I expected the output to be like that

i added : 1
i got : 1
i added : 2
i got : 2 

... and so on, but i got that output

I added : 0
I added : 1
I added : 2
I added : 3
I added : 4
I added : 5

Answer 1

Because the two threads are running at once there is no guarantee or control at all as to which runs when or in what order.

The get could run completely before the set, the set could run completely first, or they could run all mixed up.

An even interleaving like you display above is very unlikely to happen.

In that case the list has been filled fully by one thread, however the empty thread ran first and exited when it discovered there was nothing in the list.

Add diagnostics to show when each thread starts and ends and you should see that. The results will vary with every run though.

At the moment your notify does nothing as you have nothing waiting. To make the threads alternate then after each write or read you should notify and then wait. You would need to synchronize the whole block though and essentially you end up with a very complicated way to get the same result as one thread running.

Answer 2

this is loaded with race conditions. Note that the Remove thread will exit immediately if there are no elements.