如何使两个线程等待并相互通知

Question

我必须创建两个Threads，它们必须以2秒的间隔从队列中轮询和对象。

然后，第一个Thread轮询和对象等待并通知第二个轮询从其队列中轮询对象。

我读了所有关于等待和通知的内容，但没有任何关系。

任何sugestions？

第一个帖子：

public class SouthThread extends Thread {

private Queue<Car> q = new LinkedList<Car>();

public void CreateQueue() {

    Scanner input = new Scanner(System.in);

    for (int i = 0; i < 2; i++) {
        Car c = new Car();
        System.out.println("Enter registration number: ");
        String regNum = input.nextLine();
        c.setRegNum(regNum);
        q.offer(c);
    }
}

public int getQueueSize() {
    return q.size();
}

@Override
public void run() {
    while (q.size() != 0)
        try {
            while (q.size() != 0) {
                synchronized (this) {
                    System.out.print("The car with registration number: ");
                    System.out.print(q.poll().getRegNum());
                    System.out
                            .println(" have passed the bridge from the south side.");
                    this.wait(2000);
                    notify();

                }
            }
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
}

}

第二个帖子：

public class NorthThread extends Thread {
private Queue<Car> q = new LinkedList<Car>();

public void CreateQueue() {

    Scanner input = new Scanner(System.in);

    for (int i = 0; i < 2; i++) {
        Car c = new Car();
        System.out.println("Enter registration number: ");
        String regNum = input.nextLine();
        c.setRegNum(regNum);
        q.offer(c);
    }
}

public int getQueueSize() {
    return q.size();
}

@Override
public void run() {
    try {
        while (q.size() != 0) {
            synchronized (this) {
                System.out.print("The car with registration number: ");
                System.out.print(q.poll().getRegNum());
                System.out
                        .println(" have passed the bridge from the north side.");
                this.wait(2000);
                notify();
            }
        }
    } catch (InterruptedException e) {
        e.printStackTrace();
    }

}

}

主线程：

public class Main {

public static void main(String[] args) throws Exception {
    // TODO Auto-generated method stub

    SouthThread tSouthThread = new SouthThread();
    NorthThread tNorthThread = new NorthThread();

    tSouthThread.CreateQueue();
    tNorthThread.CreateQueue();

    System.out.println(tSouthThread.getQueueSize());

    tSouthThread.start();
    tNorthThread.start();


}

}

Answer 1

看起来你基本上想要实现的是一个系统，它在两个独立单元之间交替控制，这样每个单元都有一些时间来处理，然后是两秒钟的等待时间（反之亦然）。

有两种主要方法可以实现此目的：

1. 使用中央控制
2. 拥有自主通信代理

第一种方法更容易一些。 在这里，您有一个中央“主”组件，负责协调谁获得处理时间并实现等待时间。 对于这种方法，两个独立单元甚至不必是线程：

public class South {
    private Queue<Car> q = new LinkedList<Car>();

    public void CreateQueue() { ... }

    public void poll() {
        System.out.print("The car with registration number: ");
        System.out.print(q.poll().getRegNum());
        System.out.println(" have passed the bridge from the South side.");
    }
}

public class North {
    private Queue<Car> q = new LinkedList<Car>();

    public void CreateQueue() { ... }

    public void poll() {
        System.out.print("The car with registration number: ");
        System.out.print(q.poll().getRegNum());
        System.out.println(" have passed the bridge from the North side.");
    }
}

// This is the "master" class
public class Main {
    public static void main(String[] args) {
        South south = new South();
        North north = new North();

        south.CreateQueue();
        north.CreateQueue();

        boolean done = false;
        while (!done) {
            try {
                Thread.sleep(2000);
            } (catch InterruptedException) { /* TODO */ }

            north.poll();

            try {
                Thread.sleep(2000);
            } (catch InterruptedException) { /* TODO */ }

            south.poll();
        }
    }
}

请注意，North和South不会从Thread继承，即它们只是普通的旧对象。 （但是，如果你的程序更复杂并且North / South只是它的一部分，你可能想让Main（！）成为一个单独的线程并将上面的while循环放在线程的run方法中，以便其余的该程序可以同时运行。）

在第二种方法中，您没有这样的中央控制组件，但是North和South都在它们自己的单独线程中运行。 然后，这要求他们通过彼此通信来协调谁被允许处理。

public class SouthThread extends Thread {
    protected Queue<Car> q = new LinkedList<Car>();
    protected North north;

    public void CreateQueue() { ... }
    public void poll() { ... }

    public void run() {
        boolean done = false;
        while (!done) {
            // wait two seconds
            try {
                Thread.sleep(2000);
            } (catch InterruptedException) { /* TODO */ }

            // process one element from the queue
            poll();

            // notify the other thread
            synchronized (north) {
                north.notifyAll();
            }

            // wait until the other thread notifies this one
            try {
                synchronized (this) {
                    wait();
                }
            } (catch InterruptedException) { /* TODO */ }
        }
    }
}

public class NorthThread extends Thread {
    protected Queue<Car> q = new LinkedList<Car>();
    protected South south;

    public void CreateQueue() { ... }
    public void poll() { ... }

    public void run() {
        boolean done = false;
        while (!done) {
            // wait two seconds
            try {
                Thread.sleep(2000);
            } (catch InterruptedException) { /* TODO */ }

            // process one element from the queue
            poll();

            // notify the other thread
            synchronized (south) {
                south.notifyAll();
            }

            // wait until the other thread notifies this one
            try {
                synchronized (this) {
                    wait();
                }
            } (catch InterruptedException) { /* TODO */ }
        }
    }
}

public class Main {
    public static void main(String[] args) throws Exception {
        SouthThread tSouthThread = new SouthThread();
        NorthThread tNorthThread = new NorthThread();

        tSouthThread.north = tNorthThread;
        tNorthThread.south = tSouthThread;

        tSouthThread.CreateQueue();
        tNorthThread.CreateQueue();

        tSouthThread.start();
        tNorthThread.start();    
    }
}

更一般的说法：由于北方和南方似乎基本相同，因此可能不需要在两个单独的类中实现它们。 相反，只有一个实现所需功能的类并将其实例化两次就足够了：

// We can combine the functionality of North and South 
// in a single class
public class NorthSouth {
    public void CreateQueue() { ... }
    public void poll() { ... }

    // etc.
}

public class Main {
    public static void main(String[] args) {
        NorthSouth north = new NorthSouth();
        NorthSouth south = new NorthSouth();

        north.CreateQueue();
        south.CreateQueue();

        // etc.
    }
}

Answer 2

wait和notify必须引用相同的锁：当你调用object.wait(2000) ，你所说的是“我要在这里等待2000毫秒，或者直到别人调用object.notify() ，其中object是指对我说 “

我仍然不完全理解你想要实现的目标，但如果你只是想要同时执行的两个线程：

1. 做一点事
2. 等2秒
3. GOTO 1

那么你根本不需要wait / notify，你可以使用Thread.sleep()或者可能是两个java.util.Timer实例。

但同样，我不确定我是否理解正确。 :-(