一尘不染

Java使用队列的生产者/消费者线程

java

我想创建某种Producer/Consumer线程应用程序。但是我不确定在两者之间实现队列的最佳方法是什么。

因此,我提出了两个想法(这两个想法可能都是完全错误的)。我想知道哪种更好,如果它们都烂了,那么实现队列的最佳方法是什么。我关心的主要是这些示例中队列的实现。我正在扩展一个内部类的Queue类,它是线程安全的。下面是两个示例,每个示例有4个类。

主班

public class SomeApp
{
    private Consumer consumer;
    private Producer producer;

    public static void main (String args[])
    {
        consumer = new Consumer();
        producer = new Producer();
    }
} 

消费阶层

public class Consumer implements Runnable
{
    public Consumer()
    {
        Thread consumer = new Thread(this);
        consumer.start();
    }

    public void run()
    {
        while(true)
        {
            //get an object off the queue
            Object object = QueueHandler.dequeue();
            //do some stuff with the object
        }
    }
}

生产者类别

public class Producer implements Runnable
{
    public Producer()
    {
        Thread producer = new Thread(this);
        producer.start();
    }

    public void run()
    {
        while(true)
        {
            //add to the queue some sort of unique object
            QueueHandler.enqueue(new Object());
        }
    }
}

队列类

public class QueueHandler
{
    //This Queue class is a thread safe (written in house) class
    public static Queue<Object> readQ = new Queue<Object>(100);

    public static void enqueue(Object object)
    {
        //do some stuff
        readQ.add(object);
    }

    public static Object dequeue()
    {
        //do some stuff
        return readQ.get();
    }
}

要么

主班

public class SomeApp
{
    Queue<Object> readQ;
    private Consumer consumer;
    private Producer producer;

    public static void main (String args[])
    {
        readQ = new Queue<Object>(100);
        consumer = new Consumer(readQ);
        producer = new Producer(readQ);
    }
} 

消费阶层

public class Consumer implements Runnable
{
    Queue<Object> queue;

    public Consumer(Queue<Object> readQ)
    {
        queue = readQ;
        Thread consumer = new Thread(this);
        consumer.start();
    }

    public void run()
    {
        while(true)
        {
            //get an object off the queue
            Object object = queue.dequeue();
            //do some stuff with the object
        }
    }
}

生产者类别

public class Producer implements Runnable
{
    Queue<Object> queue;

    public Producer(Queue<Object> readQ)
    {
        queue = readQ;
        Thread producer = new Thread(this);
        producer.start();
    }

    public void run()
    {

        while(true)
        {
            //add to the queue some sort of unique object
            queue.enqueue(new Object());
        }
    }
}

队列类

//the extended Queue class is a thread safe (written in house) class
public class QueueHandler extends Queue<Object>
{    
    public QueueHandler(int size)
    {
        super(size); //All I'm thinking about now is McDonalds.
    }

    public void enqueue(Object object)
    {
        //do some stuff
        readQ.add();
    }

    public Object dequeue()
    {
        //do some stuff
        return readQ.get();
    }
}

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2020-03-05

共1个答案

一尘不染

Java 5+具有完成此类任务所需的所有工具。你将要:

  1. 将你所有的生产者合二为一ExecutorService;
  2. 把你所有的消费者放在另一个ExecutorService;
  3. 如有必要,请使用进行两者之间的通信BlockingQueue
    我对(3)说“如有必要”,因为根据我的经验,这是不必要的步骤。你要做的就是向消费者执行者服务提交新任务。所以:
final ExecutorService producers = Executors.newFixedThreadPool(100);
final ExecutorService consumers = Executors.newFixedThreadPool(100);
while (/* has more work */) {
  producers.submit(...);
}
producers.shutdown();
producers.awaitTermination(Long.MAX_VALUE, TimeUnit.NANOSECONDS);
consumers.shutdown();
consumers.awaitTermination(Long.MAX_VALUE, TimeUnit.NANOSECONDS);

因此producers直接提交给consumers。

2020-03-05