为什么Java Concurrency In Practice清单5.18不能用锁定原子方式完成?

时间:2013-11-21 00:13:19

标签: java concurrency

在实践中的Java并发中,第106页,它说“Memoizer3易受问题影响[两个线程看到空并开始昂贵的计算],因为执行复合操作(如果不存在)使用锁定无法成为原子的支持映射。“我不明白他们为什么说使用锁定不能成为原子。这是原始代码:

package net.jcip.examples;

import java.util.*;
import java.util.concurrent.*;

/**
 * Memoizer3
 * <p/>
 * Memoizing wrapper using FutureTask
 *
 * @author Brian Goetz and Tim Peierls
 */
public class Memoizer3 <A, V> implements Computable<A, V> {
    private final Map<A, Future<V>> cache
        = new ConcurrentHashMap<A, Future<V>>();
    private final Computable<A, V> c;

    public Memoizer3(Computable<A, V> c) {
        this.c = c;
    }

    public V compute(final A arg) throws InterruptedException {
        Future<V> f = cache.get(arg);
        if (f == null) {
            Callable<V> eval = new Callable<V>() {
                public V call() throws InterruptedException {
                    return c.compute(arg);
                }
            };
            FutureTask<V> ft = new FutureTask<V>(eval);
            f = ft;
            cache.put(arg, ft);
            ft.run(); // call to c.compute happens here
        }
        try {
            return f.get();
        } catch (ExecutionException e) {
            throw LaunderThrowable.launderThrowable(e.getCause());
        }
    }
}

为什么这样的事情不起作用?

...
public V compute(final A arg) throws InterruptedException {
    Future<V> f = null;
    FutureTask<V> ft = null;
    synchronized(this){
        f = cache.get(arg);
        if (f == null) {
            Callable<V> eval = new Callable<V>() {
                public V call() throws InterruptedException {
                    return c.compute(arg);
                }
             };
             ft = new FutureTask<V>(eval);
             f = ft;
             cache.put(arg, ft);                 
        }
    }
    if (f==ft) ft.run(); // call to c.compute happens here
    ...

1 个答案:

答案 0 :(得分:1)

当然可以通过使用锁定来实现原子化,想象一下最原始的情况:你对整个函数有一个全局锁定,然后一切都是单线程的,因此是线程安全的。我认为他们意味着别的东西,或者存在普遍的误解。

使用ConcurrentHashMap的putIfAbsent方法甚至可以改善您的代码:

public V compute(final A arg) throws InterruptedException {
  Future<V> f = cache.get(arg);
  if (f == null) {
    final Callable<V> eval = new Callable<V>() {
      public V call() throws InterruptedException {
        return c.compute(arg);
      }
    };
    final FutureTask<V> ft = new FutureTask<V>(eval);
    final Future<V> previousF = cache.putIfAbsent(arg, ft);
    if (previousF == null) {
      f = ft;
      ft.run(); 
    } else {
      f = previousF; // someone else will do the compute
    } 
  }
  return f.get();
}
最后,

f将是之前添加的值或初始值,以额外创建Callable的潜在成本,但不会多次调用计算