java - 理解java多线程同步

Question

我无法理解以下代码的输出。 我的理解是输出不会有任何特定的顺序，但 PlayerX started、PlayerX 和 PlayerX dead 应该按顺序排列。而且我们应该有所有玩家应该在缓冲区日志中并且应该在最后打印。 但有时序列是 PlayerX 启动、PlayerX 死亡 然后是 PlayerX 并且这些情况下玩家名称不在缓冲区字符串中。有人可以指出我缺少什么吗？

import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;

public class Game {

    public static void main(String[] args) {
        Ball gameBall = new Ball();
        ExecutorService executor = Executors.newFixedThreadPool(5);
        Player[] players = new Player[50];
        for (int i = 0; i < players.length; i++) {
            Player playerTemp = new Player("Player" + i, gameBall);
            executor.submit(playerTemp);
            players[i] = playerTemp;
            System.out.println(players[i].getName1() + " started");
        }

        for (int i = 0; i < players.length; i++) {

            try {
                players[i].join();
                System.out.println(players[i].getName1() + " died");
            } catch (InterruptedException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }
        /*
                 * here all thread should die and following line should display
          *all player name 
                 * without any particular order
         and should be last line.
         */

        executor.shutdown();

        System.out.println(gameBall.getLog());
    }

}

...

class Player extends Thread {

    private final String name;
    private final Ball ball;

    public Player(String aName, Ball aBall) {
        name = aName;
        ball = aBall;
    }

    @Override
    public void run() {
        ball.kick(name);
    }

    /**
     * @return the name
     */
    public String getName1() {
        return name;
    }

}

...

class Ball {

    private volatile StringBuffer log;

    public Ball() {
        log = new StringBuffer();
    }

    public synchronized void kick(String aPlayerName) {

        log.append(aPlayerName + " ");
        System.out.println(aPlayerName);

    }

    public String getLog() {
        return log.toString();
    }

}

Answer 1

首先:如果您向 Player.run() 方法添加额外的 Thread.sleep(100);，您的代码的可能性将大大增加行为错误。

您对多线程的理解其实是正确的。 但是，您对 players[i].join(); 的调用没有达到预期的效果，因为您从未启动线程 players[i]。

相反，您将其提交给 ExecutorService。此 ExecutorService 通过从其现有线程之一调用其 run() 方法来执行 Player。在您的例子中，它们是执行所有玩家工作的 5 个线程。

要获得所需的结果，您有两种可能性:

1. 不要使用ExecutorService，而是直接调用start():

   for (int i = 0; i < players.length; i++) {
       Player playerTemp = new Player("Player" + i, gameBall);
       playerTemp.start();
       players[i] = playerTemp;
       System.out.println(players[i].getName1() +" started");
   }
   

2. 使用executor.awaitTermination(..)而不是Thread.join():

   executor.shutdown();
   while(!executor.isTerminated()) {
       try {
           executor.awaitTermination(1, TimeUnit.SECONDS);
       } catch (InterruptedException e) {
           e.printStackTrace();
       }
   }