java - 使用枚举类型创建的单例，线程安全问题

Question

美好的一天，我已经创建了单例:

import java.util.Arrays;
import java.util.Collections;
import java.util.LinkedList;

public enum Singleton {
    FIRST_INSTANCE;

    String[] scrabbleLetters = {
            "a","a","a","a","a","a","a","a","a","b","b","b","b","b","b","b","b","b",
            "c","c","c","c","c","c","c","c","c","d","d","d","d","d","d","d","d","d","d",
    };

    private LinkedList<String> letterList = new LinkedList<>(Arrays.asList(scrabbleLetters));

    private Object lock = new Object();

    private Singleton() {
        Collections.shuffle(letterList);
    }

    public static Singleton getInstance() {
        return FIRST_INSTANCE;
    }

    public LinkedList<String> getLetterList() {
        synchronized (lock) {

        return FIRST_INSTANCE.letterList;
        }
    }

    public LinkedList<String> getTiles(int howManyTiles) {
        synchronized (lock) {

        LinkedList<String> tilesToSend = new LinkedList<>();
        for(int i=0; i<= howManyTiles; i++) {
            tilesToSend.add(FIRST_INSTANCE.letterList.remove(0));
        }
        return tilesToSend;

        }
    }

}

我已经用这个例子测试了它的线程安全性:

import java.util.LinkedList;

public class ScrabbleTest {
    public static void main(String[] args) {
        Runnable getTiles = () -> {

            System.out.println("In thread : " +Thread.currentThread().getName());
            Singleton newInstance = Singleton.getInstance();
            System.out.println("Instance ID: " + System.identityHashCode(newInstance));
            System.out.println(newInstance.getLetterList());

            LinkedList<String> playerOneTiles = newInstance.getTiles(7);
            System.out.println("Player : " + Thread.currentThread().getName() + playerOneTiles);
            System.out.println("Got Tiles for " + Thread.currentThread().getName());
        };

        new Thread(getTiles, "First").start();
        new Thread(getTiles, "Second").start();
    }
}

执行10次后，我确信没有问题，但是当我上次运行它时，我收到了这个堆栈跟踪:

In thread : Second
In thread : First
Instance ID: 1380197535
Instance ID: 1380197535
[d, d, b, c, b, b, a, d, c, d, a, d, c, a, a, d, c, a, a, b, d, b, b, a, b, c, a, d, c, a, c, b, c, c, b, d, d]
Player : First[d, d, b, c, b, b, a, d]
Got Tiles for First
Exception in thread "Second" java.util.ConcurrentModificationException
    at java.util.LinkedList$ListItr.checkForComodification(Unknown Source)
    at java.util.LinkedList$ListItr.next(Unknown Source)
    at java.util.AbstractCollection.toString(Unknown Source)
    at java.lang.String.valueOf(Unknown Source)
    at java.io.PrintStream.println(Unknown Source)
    at ScrabbleTest.lambda$0(ScrabbleTest.java:10)
    at java.lang.Thread.run(Unknown Source)

这种异常很少发生退出，20次执行大约1次。我发现，当不允许进行对象的并发修改时，检测到并发修改的方法可能会引发 ConcurrentModificationException。在代码中，我有一个锁可以防止这种情况，也有相同的锁用于更改和检索同步块(synchronized block)的列表。我什至无法想象为什么会发生这种情况。

Answer 1

CME 与并发性的关系并不像它的名字让您想象的那么多。 CME 最常见的情况是在单线程上下文中。然而，在这种情况下，也涉及到线程。

您的问题来自于 tilesToSend.add(FIRST_INSTANCE.letterList.remove(0));，您正在其中修改 letterList，但它正在迭代同时通过println。同步在这里没有帮助，因为您必须同步比实际可能的 block 大得多的 block 。

这里的简单解决方案是在 getLetterList() 中返回列表的副本，例如

return new LinkedList<>(FIRST_INSTANCE.letterList);

这样，当 println 迭代副本时，可以通过 remove() 修改原始列表。