/**
* 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
*
* 使用wait和notify/notifyAll来实现
*
* 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
*
* @author maple
*/
//同步互斥，等待队列， 基于AQS
package yxxy.c_021;
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();
}
}
}