Question

我有以下代码：

while(slowIterator.hasNext()) {
  performLengthTask(slowIterator.next());
}

因为迭代器和任务都很慢，所以将它们放入单独的线程是有意义的。以下是Iterator包装器的快速而脏的尝试：

class AsyncIterator<T> implements Iterator<T> {
    private final BlockingQueue<T> queue = new ArrayBlockingQueue<T>(100);

    private AsyncIterator(final Iterator<T> delegate) {
      new Thread() {
        @Override
        public void run() {
          while(delegate.hasNext()) {
            queue.put(delegate.next()); // try/catch removed for brevity
          }
        }
      }.start();
    }

    @Override
    public boolean hasNext() {
      return true;
    }

    @Override
    public T next() {
        return queue.take(); // try/catch removed for brevity
    }
    // ... remove() throws UnsupportedOperationException
  }

但是这种实现缺乏对“hasNext（）”的支持。当然可以阻止hasNext（）方法阻塞，直到它知道是否返回true。我可以在我的AsyncIterator中有一个peek对象，我可以更改hasNext（）从队列中获取一个对象并让next（）返回此窥视。但是如果已经到达委托迭代器的末尾，这将导致hasNext（）无限期地阻塞。

我可以自己做线程通信，而不是使用ArrayBlockingQueue：

private static class AsyncIterator<T> implements Iterator<T> {

  private final Queue<T> queue = new LinkedList<T>();
  private boolean delegateDone = false;

  private AsyncIterator(final Iterator<T> delegate) {
    new Thread() {
      @Override
      public void run() {
        while (delegate.hasNext()) {
          final T next = delegate.next();
          synchronized (AsyncIterator.this) {
            queue.add(next);
            AsyncIterator.this.notify();
          }
        }
        synchronized (AsyncIterator.this) {
          delegateDone = true;
          AsyncIterator.this.notify();
        }
      }
    }.start();
  }

  @Override
  public boolean hasNext() {
    synchronized (this) {
      while (queue.size() == 0 && !delegateDone) {
        try {
          wait();
        } catch (InterruptedException e) {
          throw new Error(e);
        }
      }
    }
    return queue.size() > 0;
  }

  @Override
  public T next() {
    return queue.remove();
  }

  @Override
  public void remove() {
    throw new UnsupportedOperationException();
  }
}

然而，所有额外的同步，等待和通知并没有真正使代码更具可读性，并且很容易在某处隐藏竞争条件。

有更好的想法吗？

更新

是的我确实了解常见的观察者/可观察模式。但是，通常的实现不会预见到数据流的结束，它们也不是迭代器。

我特别想要一个迭代器，因为实际上上面提到的循环存在于一个外部库中，它需要一个迭代器。

Answer 1

这是一个棘手的问题，但我想这次我得到了正确答案。（我删除了我的第一个答案。）

答案是使用哨兵。我没有测试过这段代码，为了清楚起见，我删除了try / catches：

public class AsyncIterator<T> implements Iterator<T> {

    private BlockingQueue<T> queue = new ArrayBlockingQueue<T>(100);
    private T sentinel = (T) new Object();
    private T next;

    private AsyncIterator(final Iterator<T> delegate) {
        new Thread() {
            @Override
            public void run() {
                while (delegate.hasNext()) {
                    queue.put(delegate.next());
                }
                queue.put(sentinel);
            }
        }.start();
    }

    @Override
    public boolean hasNext() {
        if (next != null) {
            return true;
        }
        next = queue.take(); // blocks if necessary
        if (next == sentinel) {
            return false;
        }
        return true;
    }

    @Override
    public T next() {
        T tmp = next;
        next = null;
        return tmp;
    }

}

这里的见解是hasNext（）需要阻止，直到下一个项目准备好。它还需要某种退出条件，并且由于线程问题，它不能使用空队列或布尔标志。哨兵在没有任何锁定或同步的情况下解决问题。

编辑：缓存“下一步”，因此可以多次调用hasNext（）。

Answer 2

或者让自己避免头痛并使用RxJava：

import java.util.Iterator;

import rx.Observable;
import rx.Scheduler;
import rx.observables.BlockingObservable;
import rx.schedulers.Schedulers;

public class RxAsyncIteratorExample {

    public static void main(String[] args) throws InterruptedException {
        final Iterator<Integer> slowIterator = new SlowIntegerIterator(3, 7300);

        // the scheduler you use here will depend on what behaviour you
        // want but io is probably what you want
        Iterator<Integer> async = asyncIterator(slowIterator, Schedulers.io());
        while (async.hasNext()) {
            performLengthTask(async.next());
        }
    }

    public static <T> Iterator<T> asyncIterator(
            final Iterator<T> slowIterator,
            Scheduler scheduler) {

        final Observable<T> tObservable = Observable.from(new Iterable<T>() {
            @Override
            public Iterator<T> iterator() {
                return slowIterator;
            }
        }).subscribeOn(scheduler);

        return BlockingObservable.from(tObservable).getIterator();
    }

    /**
     * Uninteresting implementations...
     */
    public static void performLengthTask(Integer integer)
            throws InterruptedException {
        log("Running task for " + integer);
        Thread.sleep(10000l);
        log("Finished task for " + integer);
    }

    private static class SlowIntegerIterator implements Iterator<Integer> {
        private int count;
        private final long delay;

        public SlowIntegerIterator(int count, long delay) {
            this.count = count;
            this.delay = delay;
        }

        @Override
        public boolean hasNext() {
            return count > 0;
        }

        @Override
        public Integer next() {
            try {
                log("Starting long production " + count);
                Thread.sleep(delay);
                log("Finished long production " + count);
            }
            catch (InterruptedException e) {
                throw new IllegalStateException(e);
            }
            return count--;
        }

        @Override
        public void remove() {
            throw new UnsupportedOperationException();
        }
    }

    private static final long startTime = System.currentTimeMillis();

    private static void log(String s) {
        double time = ((System.currentTimeMillis() - startTime) / 1000d);
        System.out.println(time + ": " + s);
    }
}

给我：

0.031: Starting long production 3
7.332: Finished long production 3
7.332: Starting long production 2
7.333: Running task for 3
14.633: Finished long production 2
14.633: Starting long production 1
17.333: Finished task for 3
17.333: Running task for 2
21.934: Finished long production 1
27.334: Finished task for 2
27.334: Running task for 1
37.335: Finished task for 1

异步迭代器

更新

2 个答案: