多线程解决生产者/消费者问题
这是本学期的操作系统课程的一个上机实验，用多线程来解决生产者消费者问题，源代码如下，我是初学者，尽量写的浅显些，尽量把注释写的完善些，如果有什么看不懂的可以留言，大家一起学习。
public class Semaphore {int value;
public Semaphore(int v){this.value = v;}
//定义P原语操作public synchronized void p(){
value = value-1;
if (value < 0){
try{this.wait();}
catch (InterruptedException e){}
}
}
//定义V原语操作public synchronized void v(){
value = value+1;
if (value <= 0){
this.notify();}
}
}
public class Producer extends Thread{
private Semaphore[] s;
private int i;
public Producer(int i,Semaphore[] sp){
this.i = i;
this.s = sp;
}
public void run() {while(true){
s[1].p();s[0].p();System.out.println("produce "+i++);
try {Thread.sleep(30);
}
catch (InterruptedException e) {}s[0].v();s[2].v();}
}
}
public class Consumer extends Thread {
private Semaphore[] s;
private int i;
public Consumer(int i,Semaphore[] sp){
this.i = i;
this.s = sp;
}
public void run() {while(true){
s[2].p();s[0].p();System.out.println("consumer "+i++);
try {Thread.sleep((long)(Math.random()*100));}
catch (InterruptedException e) {}s[0].v();s[1].v();}
}
}
public class pc {public static void main(String[] args) {int mutex = 1;//互斥的信号量，生产者消费者不能同时访问缓冲区
int empty = 3;//表示空闲的缓冲区的数目，初始值为3
int full = 0;//full表示有数据的缓冲区的数目，初始值为0
Semaphore[] s = new Semaphore[3];
s[0] = new Semaphore(mutex);s[1] = new Semaphore(empty);s[2] = new Semaphore(full);Producer p = new Producer(0,s);Consumer c = new Consumer(0,s);Thread pp = new Thread(p);Thread cc = new Thread(c);Thread pp1 = new Thread(p);Thread cc1 = new Thread(c);cc1.start();
pp1.start();
pp.start();
cc.start();
}
}
