java - 2个线程间的通信

Question

我想在两个线程之间建立通信，如下所示:第一个线程(Sender)向第二个线程(Receiver)发送一个整数，Receiver 显示该整数的平方。这是我的代码:Carre.java(主):

public class Carre {
    static Boolean msgArrived = Boolean.FALSE ; 
    static int n ;

    public static void main(String[] args) {
        Thread sender = new Thread(new Sender()) ;
        Thread receiver = new Thread(new Receiver()) ;
        sender.start() ;
        receiver.start() ;
    }

    public int getN() {
        return n;
    }
}

Sender.java:

import java.util.Random;

public class Sender implements Runnable {

    @Override
    public void run() {
        while(Carre.msgArrived == Boolean.TRUE) {
            try {
                wait() ;
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
        for(int i = 0; i < 10; i ++) {
            Random ran = new Random() ; 
            Carre.n = ran.nextInt(100) ;
            Carre.msgArrived = Boolean.TRUE ;
            notifyAll() ;
        }
    }
}

Receiver.java

公共(public)类 Receiver 实现 Runnable {

    @Override
    public void run() {
        while(Carre.msgArrived == Boolean.FALSE) {
            try {
                wait() ;
            } 
            catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
        System.out.println(Carre.n * Carre.n) ;
        Carre.msgArrived = Boolean.TRUE ;
        notifyAll() ;
    }
}

& 当我尝试执行代码时，我收到此错误消息:

线程“Thread-1”中的异常 java.lang.IllegalMonitorStateException

at java.lang.Object.wait(Native Method)

at java.lang.Object.wait(Object.java:502)

at Receiver.run(Receiver.java:12)

at java.lang.Thread.run(Thread.java:636)

线程“Thread-0”中的异常 java.lang.IllegalMonitorStateException

at java.lang.Object.notifyAll(Native Method)

at Sender.run(Sender.java:19)

at java.lang.Thread.run(Thread.java:636)

Answer 1

在我看来，最好不要直接使用线程，因为它们是太低级别的并发实体，因此更难使用。您应该能够通过 java.util.concurrent 更轻松地计算出您的生产者/消费者场景。封装设施。

您可以在这里查看:Producer/Consumer threads using a Queue