Java -使用具有线程间通信的两个线程打印数字序列

Question

我想用下面的格式打印数字。这应该由两个线程t1，t2来处理。有没有人能帮我增强我写的下面的代码？

First t1 should print 0-4
Then t2 should print 5-9
Then again t1 should print 10-14
Then t2 should print 15-19

0 1 2 3 4

5 6 7 8 9

10 11 12 13 14

class PrintNumber implements Runnable{
    String name;

    public void run(){
            print();
    }

    synchronized public void print(){
            for(int i=0;i< 5;i++){
                System.out.println(i+" -- "+Thread.currentThread());
            }
    }

    public static void main(String[] args){
        Runnable r = new PrintNumber();
        Thread t1 = new Thread(r,"t1");
        Thread t2 = new Thread(r,"t2");
        t1.start();
        t2.start();
    }
}

Answer 1

您可以使用两个Sempaphores，而不是使用低级wait()和notify()。每个Runnable都有一个等待的Semaphore和一个用来通知下一个的a。

import java.util.concurrent.Semaphore;

class PrintNumber implements Runnable{

    static volatile int nextStartIdx; 

    private Semaphore waitForSemaphore;

    private Semaphore next;


    public PrintNumber(Semaphore waitFor, Semaphore next) {
        this.waitForSemaphore = waitFor; 
        this.next = next;
    }


    public void run(){
        while (true) {
            print();
        }

    }

    public void print() {
        try {
            waitForSemaphore.acquire();
            int start = nextStartIdx;

            for(int i=0;i< 5;i++){
                System.out.println(String.format("%d -- %s", i + start, Thread.currentThread().getName()));
            }           

            nextStartIdx += 5;
            next.release();

        } catch (InterruptedException ie) {
            Thread.currentThread().interrupt();
        }
    }

    public static void main(String[] args){

        Semaphore a = new Semaphore(1);
        Semaphore b = new Semaphore(0);
        Thread t1 = new Thread(new PrintNumber(a,b),"t1");
        Thread t2 = new Thread(new PrintNumber(b,a),"t2");
        t1.start();
        t2.start();
    }
}

Answer 2

您可以使用等待和通知来实现sczenario的线程间通信。

class PrintNumber implements Runnable {
    String name;
    Integer count = 0;

    @Override
    public void run() {
        try {
            print();
        } catch (final InterruptedException e) {
            e.printStackTrace();
        }
    }

    synchronized public void print() throws InterruptedException {
        while (count < 15) {
            for (int i = 0; i < 5; i++) {
                count++;
                System.out.println(count + " -- " + Thread.currentThread());
            }
            notifyAll();
            wait();
        }
    }

    public static void main(final String[] args) {
        final Runnable r = new PrintNumber();
        final Thread t1 = new Thread(r, "t1");
        final Thread t2 = new Thread(r, "t2");
        t1.start();
        t2.start();
    }
}

有关详细信息，请参阅：

A simple scenario using wait() and notify() in java