码迷,mamicode.com
首页 > 其他好文 > 详细

生产者消费者(练习题:容器)

时间:2019-04-16 09:20:09      阅读:238      评论:0      收藏:0      [点我收藏+]

标签:cond   amp   his   span   getname   container   final   static   tst   

  题目(生产者消费者模式):自定义同步容器,容器容量上限为10。可以在多线程中应用,并保证数据线程安全。

方式一:synchronized

/**
 * 生产者消费者
 * wait&notify
 * wait/notify都是和while配合应用的。可以避免多线程并发判断逻辑失效问题。
 */
import java.util.LinkedList;
import java.util.concurrent.TimeUnit;

public class TestContainer01<E> {

    private final LinkedList<E> list = new LinkedList<>();
    private final int MAX = 10;
    private int count = 0;

    public static void main(String[] args) {
        final TestContainer01<String> c = new TestContainer01<>();
        for (int i = 0; i < 10; i++) {
            new Thread(new Runnable() {
                @Override
                public void run() {
                    for (int j = 0; j < 5; j++) {
                        System.out.println(c.get());
                    }
                }
            }, "consumer" + i).start();
        }
        try {
            TimeUnit.SECONDS.sleep(2);
        } catch (InterruptedException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        for (int i = 0; i < 2; i++) {
            new Thread(new Runnable() {
                @Override
                public void run() {
                    for (int j = 0; j < 25; j++) {
                        c.put("container value " + j);
                    }
                }
            }, "producer" + i).start();
        }
    }

    public synchronized int getCount() {
        return count;
    }

    public synchronized void put(E e) {
        while (list.size() == MAX) {
            try {
                this.wait();
            } catch (InterruptedException e1) {
                e1.printStackTrace();
            }
        }

        list.add(e);
        count++;
        this.notifyAll();
    }

    public synchronized E get() {
        E e = null;
        while (list.size() == 0) {
            try {
                this.wait();
            } catch (InterruptedException e1) {
                e1.printStackTrace();
            }
        }
        e = list.removeFirst();
        count--;
        this.notifyAll();
        return e;
    }

}

方法二:ReentrantLock

/**
 * 生产者消费者
 * 重入锁&条件
 * 条件 - Condition, 为Lock增加条件。当条件满足时,做什么事情,如加锁或解锁。如等待或唤醒
 */

import java.util.LinkedList;
import java.util.concurrent.TimeUnit;
import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;

public class TestContainer02<E> {

    private final LinkedList<E> list = new LinkedList<>();
    private final int MAX = 10;
    private int count = 0;

    private Lock lock = new ReentrantLock();
    private Condition producer = lock.newCondition();
    private Condition consumer = lock.newCondition();

    public static void main(String[] args) {
        final TestContainer02<String> c = new TestContainer02<>();
        for (int i = 0; i < 10; i++) {
            new Thread(new Runnable() {
                @Override
                public void run() {
                    for (int j = 0; j < 5; j++) {
                        System.out.println(c.get());
                    }
                }
            }, "consumer" + i).start();
        }
        try {
            TimeUnit.SECONDS.sleep(2);
        } catch (InterruptedException e1) {
            e1.printStackTrace();
        }
        for (int i = 0; i < 2; i++) {
            new Thread(new Runnable() {
                @Override
                public void run() {
                    for (int j = 0; j < 25; j++) {
                        c.put("container value " + j);
                    }
                }
            }, "producer" + i).start();
        }
    }

    public int getCount() {
        return count;
    }

    public void put(E e) {
        lock.lock();
        try {
            while (list.size() == MAX) {
                System.out.println(Thread.currentThread().getName() + " 等待。。。");
                // 进入等待队列。释放锁标记。
                // 借助条件,进入的等待队列。
                producer.await();
            }
            System.out.println(Thread.currentThread().getName() + " put 。。。");
            list.add(e);
            count++;
            // 借助条件,唤醒所有的消费者。
            consumer.signalAll();
        } catch (InterruptedException e1) {
            e1.printStackTrace();
        } finally {
            lock.unlock();
        }
    }

    public E get() {
        E e = null;
        lock.lock();
        try {
            while (list.size() == 0) {
                System.out.println(Thread.currentThread().getName() + " 等待。。。");
                // 借助条件,消费者进入等待队列
                consumer.await();
            }
            System.out.println(Thread.currentThread().getName() + " get 。。。");
            e = list.removeFirst();
            count--;
            // 借助条件,唤醒所有的生产者
            producer.signalAll();
        } catch (InterruptedException e1) {
            e1.printStackTrace();
        } finally {
            lock.unlock();
        }

        return e;
    }

}

    区别:synchronized的notify会唤醒所有的线程(生产者+消费者),而ReentrantLock的Condition会精准的唤醒生产者或者消费者。但是两者效率差别不大。

生产者消费者(练习题:容器)

标签:cond   amp   his   span   getname   container   final   static   tst   

原文地址:https://www.cnblogs.com/jing99/p/10714420.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!