Question

我使用wait和notify方法进行研究以进行同步。

我写了一些小例子并且不明白为什么我看到这个输出：

take 1
release 1
release 2

我的代码：

主：

public class CustomSemaphoreTest {
    public static void main(String[] args) throws InterruptedException {

        Semaphore semaphore = new Semaphore();
        SendingThread sender = new SendingThread(semaphore);
        RecevingThread receiver = new RecevingThread(semaphore);
        sender.start();
        Thread.sleep(300);
        receiver.start();

    }
}

发件人：

class SendingThread extends Thread {
    volatile Semaphore semaphore = null;
    int count = 1;

    public SendingThread(Semaphore semaphore) {
        this.semaphore = semaphore;
    }

    public void run() {
        this.semaphore.take();
    }
}

接收器：

class RecevingThread extends Thread {
    volatile Semaphore semaphore = null;
    int count = 1;

    public RecevingThread(Semaphore semaphore) {
        this.semaphore = semaphore;
    }

    public void run() {
        while (true) {
            try {
                this.semaphore.release();
                Thread.sleep(20);
            } catch (InterruptedException e) {
            }
            // System.out.println("get signal " + count++);
        }
    }
}

旗语：

class Semaphore {
    private boolean signal = false;
    int rCount = 1;
    int sCount = 1;

    public synchronized void take() {
        System.out.println("take " + sCount++);
        this.signal = true;
        this.notify();
        try {
            wait();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }

    public synchronized void release() throws InterruptedException {
        System.out.println("release " + rCount++);
        notify();
        while (!this.signal) {
            wait();
        }
        this.signal = false;
    }
}

我是如何想象执行这个程序的？：

时间0：thread1：sender.start() - ＆gt; ...-＆gt;

public synchronized void take() {
        System.out.println("take " + sCount++);
        this.signal = true;
        this.notify(); //try wake up waiting threads but noone sleep thus have not affect
        try {
            wait(); // release monitor and await notify
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }

因此，我在屏幕take 1

中看到了

时间300：主题2 receiver.start(); - ＆gt;

 public synchronized void release() throws InterruptedException {
        System.out.println("release " + rCount++);  //section is rid thus this lines outputes
        notify(); // notify thread1
        while (!this.signal) {
            wait(); // pass the control to thread1 and await notification .....
        }
        this.signal = false;
    }

因此我没有期望看到release 2

请澄清我的误解。

Answer 1

在 signal = false之后设置wait() 。

事件顺序。

sender来电take()

sender打印

signal设置为true

receiver已收到通知

sender来电wait()

receiver醒来

receiver打印

sender已收到通知

signal是true所以没有wait()

signal设置为false

release()方法结束

receiver循环

receiver打印

signal false wait()

sender立即醒来并立即退出

所以你有几个问题

您的sender无法循环

您在错误的时间设置signal

使用wait和notify执行代码的顺序

1 个答案: