Java如何避免在循环中使用thread.sleep（）

Question

从我的main开始，我将启动两个线程，称为producer和consumer。两者都包含while(true)循环。生产者循环是UDP服务器，因此不需要Hibernate。我的问题出在消费者方面。使用者循环将对象从链接队列中移除，并将其传递给一个函数进行进一步处理。根据研究，在循环中使用线程Hibernate不是一个好的实践，因为有时O/S在设定时间结束时不会释放。如果我删除线程Hibernate，当应用程序是理想的，它拖到20%到30%的CPU。

class Producer implements Runnable {
    private DatagramSocket dsocket;
    FError fer = new FError();

    int port =1548;
    ConcurrentLinkedQueue<String> queue;

    Producer(ConcurrentLinkedQueue<String> queue){
        this.queue = queue; 
    }

    @Override
    public void run() {

        try {

            // Create a socket to listen on the port.
            dsocket = new DatagramSocket(port);
            // Create a buffer to read datagrams into.
            byte[] buffer = new byte[30000];
            // Create a packet to receive data into the buffer
            DatagramPacket packet = new DatagramPacket(buffer,
            buffer.length);

            while (true) {
                try {

                   // Wait to receive a datagram
                    dsocket.receive(packet);
                    //Convert the contents to a string,
                    String msg = new String(buffer, 0, packet.getLength());

                    int ltr = msg.length();
                     // System.out.println("MSG =" + msg);

                    if(ltr>4)
                    {

                        SimpleDateFormat sdfDate = new SimpleDateFormat  ("yyyy-MM-dd HH:mm:ss");//dd/MM/yyyy

                        Date now = new Date();
                        String strDate = sdfDate.format(now);

                        //System.out.println(strDate);

                        queue.add(msg + "&" + strDate);

                     // System.out.println("MSG =" + msg);
                    }

                  // Reset the length of the packet before reusing it.
                   packet.setLength(buffer.length);

                } catch (IOException e) {
                    fer.felog("svr class", "producer", "producer thread",e.getClass().getName() + ": " + e.getMessage());
                    dsocket.close();
                    break; 
                }
            }

        } catch (SocketException e) {
          fer.felog("svr class", "producer","Another App using the udp port " + port, e.getClass().getName() + ": " + e.getMessage()); 

        }

    }

}

class Consumer implements Runnable {

    String str;  
    ConcurrentLinkedQueue<String> queue;

    Consumer(ConcurrentLinkedQueue<String> queue) {
        this.queue = queue;  
    }

    @Override
    public void run() {

        while (true) {
            try {

                while ((str = queue.poll()) != null) {

                    call(str);  // do further processing

                   }
            } catch (IOException e) {
                ferpt.felog("svr class", "consumer", "consumer thread", e.getClass().getName() + ": " + e.getMessage());
                break;
            }

            try {
                Thread.sleep(500);
            } catch (InterruptedException ex) {

                ferpt.felog("svr class", "consumer","sleep", ex.getClass().getName() + ": " + ex.getMessage());
            }

        }

    }

}

Answer 1

解决问题的正确方法是使用阻塞队列。它为您提供了几个优点：

• 不会浪费cpu忙于等待
• 可以有有限的容量--假设您有一个快速的生产者，但一个缓慢的使用者->如果队列的大小不受限制，那么您的应用程序可以很容易地达到OutOfMemory条件

这里有一个小演示，你可以玩一下：

import java.util.concurrent.ArrayBlockingQueue;
import java.util.concurrent.BlockingQueue;

public class ProdConsTest {
    public static void main(String[] args) throws InterruptedException {
        final BlockingQueue<Integer> queue = new ArrayBlockingQueue<>(10);
        final Runnable producer = () -> {
            for (int i = 0; i < 1000; i++) {
                try {
                    System.out.println("Producing: " + i);
                    queue.put(i);

                    //Adjust production speed by modifying the sleep time
                    Thread.sleep(100);
                } catch (InterruptedException e) {
                    //someone signaled us to terminate
                    break;
                }
            }
        };

        final Runnable consumer = () -> {
            while (true) {
                final Integer integer;
                try {
                    //Uncomment to simulate slow consumer:
                    //Thread.sleep(1000);

                    integer = queue.take();
                } catch (InterruptedException e) {
                    //someone signaled us to terminate
                    break;
                }
                System.out.println("Consumed: " + integer);
            }
        };


        final Thread consumerThread = new Thread(consumer);
        consumerThread.start();

        final Thread producerThread = new Thread(producer);
        producerThread.start();

        producerThread.join();
        consumerThread.interrupt();
        consumerThread.join();
    }
}

现在取消对使用者中的sleep()的注释，并观察应用程序的优点。如果您使用的是基于计时器的解决方案，如建议的ScheduleDexecutorService或者您正忙于等待，那么使用fast producer，队列将无法控制地增长，并最终使应用程序崩溃

Answer 2

您可以更改代码以合并ScheduleDexecutorService，而不是使使用者扩展为可运行的，该ScheduleDexecutorService每半秒运行一次队列轮询，而不是使线程休眠。这方面的一个例子是

public void schedule() {
    ScheduledExecutorService executor = Executors.newSingleThreadScheduledExecutor();
    executor.scheduleAtFixedRate(() -> {
        String str;
        try {
            while ((str = queue.poll()) != null) {
                call(str);  // do further processing
            }
        } catch (IOException e) {
            ferpt.felog("svr class", "consumer", "consumer thread", e.getClass().getName() + ": " + e.getMessage());
        }
    }, 0, 500, TimeUnit.MILLISECONDS);
}