如何知道哪个线程首先从两个线程中完成执行 [英] How to know which thread completes execution first out of two threads

问题描述

我有两个线程A和B.如果A首先完成，那么我必须执行function1 else如果B先完成，我需要执行函数2.我知道这两个线程中哪一个先完成了执行？ / p>

I have two threads A and B. If A completes first, then I have to execute function1 else if B completes first, I need to execute function 2. HOw do I know which of the two threads has completed execution first?

推荐答案

import java.util.concurrent.Semaphore;

public class ThreadTest {
    private Semaphore semaphore = new Semaphore(0);
    private String winner;

    private synchronized void finished(String threadName) {
        if (winner == null) {
            winner = threadName;
        }
        semaphore.release();
    }

    public void run() {
        Runnable r1 = new Runnable() {
            public void run() {
                try {
                    Thread.sleep((long) (5000 * Math.random()));
                }
                catch (InterruptedException e) {
                    // ignore
                }
                finally {
                    finished("thread 1");
                }
            }
        };

        Runnable r2 = new Runnable() {
            public void run() {
                try {
                    Thread.sleep((long) (5000 * Math.random()));
                }
                catch (InterruptedException e) {
                    // ignore
                }
                finally {
                    finished("thread 2");
                }
            }
        };

        Thread t1 = new Thread(r1);
        Thread t2 = new Thread(r2);
        t1.start();
        t2.start();
        try {
            semaphore.acquire();
            System.out.println("The winner is " + winner);
        }
        catch (InterruptedException e) {
            System.out.println("No winner");
            Thread.currentThread().interrupt();
        }
    }

    public static void main(String[] args) {
        new ThreadTest().run();
    }
}

此解决方案的优势在于尽快获胜当第一个线程结束时，而不是必须等到所有线程都完成。

This solution has the advantage of having a winner as soon as the first thread finishes, instead of having to wait until all the threads are done.

编辑：

Aa， CountDownLatch 是这个问题的更好的抽象。因此可以写出相同的算法：

Aa indicated by jtahlborn, CountDownLatch is a better abstraction for this problem. The same algorithm can thus be written as this :

import java.util.concurrent.CountDownLatch;

public class ThreadTest {
    private CountDownLatch latch = new CountDownLatch(1);
    private String winner;

    private synchronized void finished(String threadName) {
        if (winner == null) {
            winner = threadName;
        }
        latch.countDown();
    }

    public void run() {
        Runnable r1 = new Runnable() {
            public void run() {
                try {
                    Thread.sleep((long) (5000 * Math.random()));
                }
                catch (InterruptedException e) {
                    // ignore
                }
                finally {
                    finished("thread 1");
                }
            }
        };

        Runnable r2 = new Runnable() {
            public void run() {
                try {
                    Thread.sleep((long) (5000 * Math.random()));
                }
                catch (InterruptedException e) {
                    // ignore
                }
                finally {
                    finished("thread 2");
                }
            }
        };

        Thread t1 = new Thread(r1);
        Thread t2 = new Thread(r2);
        t1.start();
        t2.start();
        try {
            latch.await();
            System.out.println("The winner is " + winner);
        }
        catch (InterruptedException e) {
            System.out.println("No winner");
            Thread.currentThread().interrupt();
        }
    }

    public static void main(String[] args) {
        new ThreadTest().run();
    }
}

如果你想等待两个线程完成在显示获胜者之前，您只需将锁存器初始化为2而不是1。

If you want to wait for the two threads to finish before displaying the winner, you just have to initialize the latch to 2 instead of 1.

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