Question

美好的一天，我创造了Singleton：

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;

        }
    }

}

我用这个例子测试了线程的安全性：

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次之后，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)

此异常很少退出，大约1次执行20次。我发现，当不允许这样的修改时，检测到对象的并发修改的方法可能抛出ConcurrentModificationException。在代码中我有一个锁可以防止这种情况，有相同的锁用于更改和检索同步块的列表。我甚至没想到为什么会这样。

Answer 1

CME与并发性没有多大关系，因为名称可能会让你思考。 CME最常见的情况是单线程上下文。但是，在这种情况下，也涉及线程。

您的问题来自tilesToSend.add(FIRST_INSTANCE.letterList.remove(0));，其中您正在修改letterList，但println会同时对其进行迭代。同步在这里不会有所帮助，因为您必须同步比实际可能更大的块。

这里的简单解决方案是在getLetterList()中返回列表的副本，如

return new LinkedList<>(FIRST_INSTANCE.letterList);

这样，remove()可以修改原始列表，而println正在迭代副本。

使用枚举类型创建单例，线程安全问题

1 个答案: