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我正在用多线程和 Java 并发特性做一些练习。我有 1 个生产者和 4 个消费者。现在我的问题是：当我确定生产者已经在 BlockingQueue 中完成生产时，还有其他更聪明的方法来阻止消费者吗？现在我在队列中使用 -1 Integer，但看起来非常初级！谢谢

public class Exercise {

static class Producer implements Runnable {
    int counter=0;
    private BlockingQueue<Integer> queue;
    Producer(BlockingQueue<Integer> q) { 
        queue = q;
    }
    public void run() {
        try {
            while (counter<100000000) {
                queue.put(produce());
            }
            queue.put(new Integer(-1));
            queue.put(new Integer(-1));
            queue.put(new Integer(-1));
            queue.put(new Integer(-1));
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }

    Integer produce() {
        counter++;
        return new Integer(counter);

    }
}

static class Consumer implements Runnable {
    private final BlockingQueue<Integer> queue;
    private String name;
    private long sum;

    Consumer(BlockingQueue<Integer> q, String name) { 
        queue = q; 
        this.name=name;
        sum=0;
    }

    public void run() {
        try {
            int x=0;
            while (x>=0) {
                x=queue.take();
                if(x!=-1)sum+=x;
            }
            System.out.println(sum+" of "+ name);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }
}



public static void main(String[] args) {

    ExecutorService exec = Executors.newFixedThreadPool(6);
    BlockingQueue<Integer> q =new LinkedTransferQueue<Integer>();
    Producer p=new Producer(q);    
    Consumer c1 = new Consumer(q,"consumer1");
    Consumer c2 = new Consumer(q,"consumer2");
    Consumer c3 = new Consumer(q,"consumer3");
    Consumer c4 = new Consumer(q,"consumer4");
    exec.submit(p);
    exec.submit(c1);
    exec.execute(c2);
    exec.submit(c3);
    exec.execute(c4);
    exec.shutdown();
}

}

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1 回答 1

1

您可以使用毒丸，但是使用这种毒丸的更强大的方法是不将其从队列中删除（或者如果您这样做则将其放回）这种方式生产者不需要知道您有多少消费者有。

顺便说一句，我会避免使用显式装箱，因为它更冗长且速度较慢。

代替

queue.put(new Integer(-1));

你可以写

queue.put(-1);

甚至更好

static final int POISON_PILL = -1;

// on the producer
queue.put(POISON_PILL);

// on the consumer
while ((x = queue.take()) != POISON_PILL) {
    sum += x;
queue.put(POISON_PILL);
于 2015-02-04T09:46:48.623 回答