如何知道哪个线程首先从两个线程中完成执行

我有两个线程A和B.如果A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执行？

Answer 1

您可以使用以下内容，只有在前一个值为null时才会设置。 （即使你只有一个线程来确保一旦设置值不改变，也可以使用它）

AtomicReference ref = new AtomicReference(); ref.compareAndSet(null, value); 

Answer 2

（许多）可能的解决方案之一。 有一些共享标志变量

 import java.util.concurrent.atomic.AtomicInteger; final AtomicInteger winner = new AtomicInteger(0); 

然后在第一个线程中调用thread的run（）方法结束

 winner.compareAndSet(0, 1); 

在第二个线程

 winner.compareAndSet(0, 2); 

这种方式primefaces整数只允许在首先调用compareAndSet的线程中设置非零值。

然后你可以获得执行结果作为胜利者指数

 winner.get() 

在生产代码中，我建议使用一些常量来表示初始值和线程索引。

Answer 3

 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(); } } 

该解决方案的优点是，一旦第一个线程完成就有赢家，而不必等到所有线程都完成。

编辑：

由jtahlborn指出， CountDownLatch是这个问题的更好的抽象。 因此可以编写相同的算法：

 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(); } } 

如果要在显示获胜者之前等待两个线程完成，则只需将锁存器初始化为2而不是1。