按顺序java运行3个线程

Question

我有3个线程第1次打印第2次打印B第3次打印C.

我想按顺序打印ABCABCABC等等......

所以我写了下面的程序,但我无法实现同样的目标.我知道当status = 1时,例如B1和C1线程正在等待,当我执行notifyAll()时,两个等待线程唤醒并根据CPU分配,它可能会打印B或C.

在这种情况下,我希望在A之后只打印B.

我需要做什么修改.

public class NotifyAllExample {

    int status=1;
    public static void main(String[] args) {

        NotifyAllExample notifyAllExample = new NotifyAllExample();

        A1 a=new A1(notifyAllExample);
        B1 b=new B1(notifyAllExample);
        C1 c=new C1(notifyAllExample);

        a.start();
        b.start();
        c.start();
    }
}

class A1 extends Thread{
    NotifyAllExample notifyAllExample;

    A1(NotifyAllExample notifyAllExample){
        this.notifyAllExample = notifyAllExample;
    }

    @Override
    public void run() {

        try{
            synchronized (notifyAllExample) {

                for (int i = 0; i < 100; i++) {

                    if(notifyAllExample.status!=1){
                        notifyAllExample.wait();
                    }

                    System.out.print("A ");
                    notifyAllExample.status = 2;
                    notifyAllExample.notifyAll();
                }

            }
        }catch (Exception e) {
            System.out.println("Exception 1 :"+e.getMessage());
        }

    }

}

class B1 extends Thread{

    NotifyAllExample notifyAllExample;

    B1(NotifyAllExample notifyAllExample){
        this.notifyAllExample = notifyAllExample;
    }

    @Override
    public void run() {

        try{
            synchronized (notifyAllExample) {

                for (int i = 0; i < 100; i++) {

                    if(notifyAllExample.status!=2){
                        notifyAllExample.wait();
                    }

                    System.out.print("B ");
                    notifyAllExample.status = 3;
                    notifyAllExample.notifyAll();
                }

            }
        }catch (Exception e) {
            System.out.println("Exception 2 :"+e.getMessage());
        }

    }
}


class C1 extends Thread{

    NotifyAllExample notifyAllExample;

    C1(NotifyAllExample notifyAllExample){
        this.notifyAllExample = notifyAllExample;
    }

    @Override
    public void run() {

        try{
            synchronized (notifyAllExample) {

                for (int i = 0; i < 100; i++) {

                    if(notifyAllExample.status!=3){
                        notifyAllExample.wait();
                    }

                    System.out.print("C ");
                    notifyAllExample.status = 1;
                    notifyAllExample.notifyAll();
                }

            }
        }catch (Exception e) {
            System.out.println("Exception 3 :"+e.getMessage());
        }

    }
}

Answer 1

将这些IF语句转换为WHILE语句以获得所需的行为:

if (notifyAllExample.status != 2){
    notifyAllExample.wait();
}

至

while (notifyAllExample.status != 2){
    notifyAllExample.wait();
}

这将确保如果通知一个线程,它将不会退出while循环,直到状态值达到预期值.

另外,标记status为volatile,以便线程不具有本地副本.

Answer 2

 public class RunThreadsInOrder implements Runnable {

    static int numThread = 1;
    static int threadAllowedToRun = 1;
    int myThreadID;
    private static Object myLock = new Object();

    public RunThreadsInOrder() {
        this.myThreadID = numThread++;
        System.out.println("Thread ID:" + myThreadID);
    }

    @Override
    public void run() {
        synchronized (myLock) {
            while (myThreadID != threadAllowedToRun) {
                try {
                    myLock.wait();
                } catch (InterruptedException e) {

                } catch (Exception e) {}
            }
            try {
                Thread.sleep(2000);
            } catch (InterruptedException e) {
            }

            System.out.println("myThreadID is running: " + myThreadID);
            myLock.notifyAll();
            threadAllowedToRun++;
        }
    }

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

        Thread t1 = new Thread(new RunThreadsInOrder());
        Thread t2 = new Thread(new RunThreadsInOrder());
        Thread t3 = new Thread(new RunThreadsInOrder());
        Thread t4 = new Thread(new RunThreadsInOrder());
        Thread t5 = new Thread(new RunThreadsInOrder());
        Thread t6 = new Thread(new RunThreadsInOrder());
        Thread t7 = new Thread(new RunThreadsInOrder());

        t7.start();
        t6.start();
        t5.start();
        t4.start();
        t3.start();
        t2.start();
        t1.start();

    }
}