zoukankan      html  css  js  c++  java
  • Condition使用

    面试题:写一个固定容量同步容器,拥有put和get方法,以及getCount方法,
        能够支持2个生产者线程以及10个消费者线程的阻塞调用

    有两种方法

    1.使用wait和notify/notifyAll来实现

    import java.util.LinkedList;
    import java.util.concurrent.TimeUnit;
    
    public class MyContainer1<T> {
        final private LinkedList<T> lists = new LinkedList<>();
        final private int MAX = 10; //最多10个元素
        private int count = 0;
        
        
        public synchronized void put(T t) {
            while(lists.size() == MAX) { //想想为什么用while而不是用if?
                try {
                    this.wait(); //effective java
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
            
            lists.add(t);
            ++count;
            this.notifyAll(); //通知消费者线程进行消费
        }
        
        public synchronized T get() {
            T t = null;
            while(lists.size() == 0) {
                try {
                    this.wait();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
            t = lists.removeFirst();
            count --;
            this.notifyAll(); //通知生产者进行生产
            return t;
        }
        
        public static void main(String[] args) {
            MyContainer1<String> c = new MyContainer1<>();
            //启动消费者线程
            for(int i=0; i<10; i++) {
                new Thread(()->{
                    for(int j=0; j<5; j++) System.out.println(c.get());
                }, "c" + i).start();
            }
            
            try {
                TimeUnit.SECONDS.sleep(2);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            
            //启动生产者线程
            for(int i=0; i<2; i++) {
                new Thread(()->{
                    for(int j=0; j<25; j++) c.put(Thread.currentThread().getName() + " " + j);
                }, "p" + i).start();
            }
        }
    }

    2.使用Lock和Condition来实现
     对比两种方式,Condition的方式可以更加精确的指定哪些线程被唤醒

    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 MyContainer2<T> {
        final private LinkedList<T> lists = new LinkedList<>();
        final private int MAX = 10; //最多10个元素
        private int count = 0;
        
        private Lock lock = new ReentrantLock();
        private Condition producer = lock.newCondition();
        private Condition consumer = lock.newCondition();
        
        public void put(T t) {
            try {
                lock.lock();
                while(lists.size() == MAX) { //想想为什么用while而不是用if?
                    producer.await();
                }
                
                lists.add(t);
                ++count;
                consumer.signalAll(); //通知消费者线程进行消费
            } catch (InterruptedException e) {
                e.printStackTrace();
            } finally {
                lock.unlock();
            }
        }
        
        public T get() {
            T t = null;
            try {
                lock.lock();
                while(lists.size() == 0) {
                    consumer.await();
                }
                t = lists.removeFirst();
                count --;
                producer.signalAll(); //通知生产者进行生产
            } catch (InterruptedException e) {
                e.printStackTrace();
            } finally {
                lock.unlock();
            }
            return t;
        }
        
        public static void main(String[] args) {
            MyContainer2<String> c = new MyContainer2<>();
            //启动消费者线程
            for(int i=0; i<10; i++) {
                new Thread(()->{
                    for(int j=0; j<5; j++) System.out.println(c.get());
                }, "c" + i).start();
            }
            
            try {
                TimeUnit.SECONDS.sleep(2);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            
            //启动生产者线程
            for(int i=0; i<2; i++) {
                new Thread(()->{
                    for(int j=0; j<25; j++) c.put(Thread.currentThread().getName() + " " + j);
                }, "p" + i).start();
            }
        }
    }
  • 相关阅读:
    leetcode35. search Insert Position
    leetcode26.Remove Duplicates from Sorted Array
    leetcode46.Permutation & leetcode47.Permutation II
    leetcode.5 Longest Palindromic Substring
    [转载] C++中new和malloc的区别
    [转载] C++中的自由存储区和堆
    xCode8以及iOS10 的新特性
    cell上添加倒计时,以及时差问题的解决
    cell的复用机制
    iOS 懒加载模式
  • 原文地址:https://www.cnblogs.com/ganchuanpu/p/7842546.html
Copyright © 2011-2022 走看看