Java线程，我怎样才能让一个线程总是先停止或结束？

Question

我想让一个线程B稍后结束，这样结果总是在最后显示B9。我怎样才能做到这一点？这是我的代码。

package thread;
import java.lang.*;
import java.lang.Thread;
import java.lang.Runnable;

class Numprintt implements Runnable {
    String myName;
    public Numprintt(String name) {
        myName = name;
    }
    public void run() {
        for(int i = 0; i < 10; i++) {
            System.out.print(myName + i + " ");
        }
    }
}


public class MyRunnableTest {
    public static void main(String[] ar) {

        Numprintt A = new Numprintt("A");
        Numprintt B = new Numprintt("B");

        Thread t1 = new Thread(A);
        Thread t2 = new Thread(B);

        t1.start();
        t2.start();

        try {
            t1.join(); t2.join();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}

我的结果可能显示 A1、A2、B1、....、B9 或者， B1, B2, A1, A2, ..., A9 我希望结果始终像前一个一样显示。

Answer 1

您可以利用 CountDownLatch 的功能。 CountDownLatch 使您能够等待从其他线程到同一个 CountDownLatch 对象的倒计时。这是您的问题的基本解决方案。

import java.lang.*;
import java.lang.Thread;
import java.lang.Runnable;
import java.util.concurrent.CountDownLatch;
import java.util.concurrent.locks.ReentrantLock;

class Numprintt implements Runnable {
    private final CountDownLatch latch;
    private String myName;

    public Numprintt(String name, CountDownLatch latch) {
        this.myName = name;
        this.latch = latch;
    }

    public void run() {
        for (int i = 0; i < 10; i++) {
            if (i == 9 && latch != null) {
                try {
                    latch.await();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }

            System.out.print(myName + i + " ");
        }
    }
}


public class MyRunnableTest {

    public static void main(String[] ar) {

        CountDownLatch latch = new CountDownLatch(1);

        Numprintt A = new Numprintt("A", null);
        Numprintt B = new Numprintt("B", latch);


        Thread t1 = new Thread(A);
        Thread t2 = new Thread(B);

        t1.start();
        t2.start();

        try {
            t1.join();
            latch.countDown();
            t2.join();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
}