Java并发之:生产者消费者问题

生产者消费者问题是Java并发中的常见问题之一，在实现时，一般可以考虑使用juc包下的BlockingQueue接口，至于具体使用哪个类，则就需要根据具体的使用场景具体分析了。本文主要实现一个生产者消费者的原型，以及实现一个生产者消费者的典型使用场景。

第一个问题：实现一个生产者消费者的原型。

import java.util.concurrent.*;

class Consumer implements Runnable {
    BlockingQueue q = null;

    public Consumer(BlockingQueue q) {
        this.q = q;
    }

    @Override 
    public void run() {
        while(true) {
            try {
                q.take();
                System.out.println("Consumer has taken a product.");
            }catch(InterruptedException e) {
    
            }
        }
    }
}

class Producer implements Runnable {
    BlockingQueue q = null;
    
    public Producer(BlockingQueue q) {
        this.q = q;
    }

    @Override
    public void run() {
        while(true) {
            try { // note that if there is any chance that block, usually we need a InterruptedException
                q.put(new Object());
                System.out.println("Producer has puted a product.");
            }catch(InterruptedException e) {

            }
        }
    }


}

public class JC_ProducerConsumerPrototype {
    static int queueCapacity = 1024;
    //static BlockingQueue<Object> q = new ArrayBlockingQueue<Object>(queueCapacity); // Can also compile
    static BlockingQueue q = new ArrayBlockingQueue(queueCapacity); // ABQ must has a capacity
    public static void main(String[] args) {
        Thread t1 = new Thread(new Producer(q));
        Thread t2 = new Thread(new Consumer(q));
        t1.start();
        t2.start();
    }


}

第二个问题，现在假设生产者是在读取磁盘上的多个log文件，对于每一个文件，依次读取文件中的每一行，也就是一条log记录；消费者需要读取并分析这些记录，假设消费者是计算密集型的。如何在生产者消费者原型的基础上实现这些功能？

这个场景在server端开发中是经常碰到的，因为在Server端，不可避免地会产生大量的日志文件。

import java.util.concurrent.*;
import java.io.*;
import java.nio.*;
import java.nio.file.*;
import java.util.*;
import java.nio.charset.*;


class Producer implements Runnable {
    BlockingQueue q = null;
    String fileName = null;
    CountDownLatch latch = null;
    
    public Producer(BlockingQueue q,String fileName,CountDownLatch latch) {
        this.q = q;
        this.fileName = fileName;
        this.latch = latch;
    }

    @Override
    public void run() {
        Path path = Paths.get(".",fileName);
        try{
            List<String> lines = Files.readAllLines(path,StandardCharsets.UTF_8);
            for(int i=lines.size();i>0;i--){
                try{
                    q.put(lines.get(i));
                }catch(InterruptedException e) {

                }
            }
        }catch(IOException e){

        }
        latch.countDown();
    }
}

class Consumer implements Runnable {
    BlockingQueue<String> q = null;
    Boolean done = false;
    
    public Consumer(BlockingQueue q,Boolean done){
        this.q = q;
        this.done = done;
    }

    @Override
    public void run(){
        while(!done||q.size()!=0){
            try{
                q.take();
            }catch(InterruptedException e){

            }
        }
    }
}

public class JC_ProducerConsumerHandlingLog{
    public static int fileCount = 1024;
    public static String[] fileNames = new String[fileCount];
    public static int cpuCount = 8;
    public static CountDownLatch latch = new CountDownLatch(fileCount);
    public static volatile boolean done = false;
    public static BlockingQueue<String> q = new LinkedBlockingQueue<String>(fileCount);//one thread for one file

    public static void main(String[] args){
        for(int i=0;i<fileCount;i++){
            Thread t = new Thread(new Producer(q,fileNames[i],latch));
            t.start();
        }
        for(int i=0;i<cpuCount;i++){//for computing tasks, we don't need too many threads.
            Thread t = new Thread(new Consumer(q,done));
            t.start();
        }
        try{
            latch.await();
            done = true;
        }catch(InterruptedException e){

        }

    }
}

 需要稍微注意一下线程数的选择，对于计算密集型的任务，我认为线程数达到cpu的核数比较合理(在不考虑超线程的情况下，也就是说一个核只有一个线程)。有不同意见欢迎跟我交流！