分享:java多线程实现生产者消费者问题

//生产者消费者问题
//生产者线程
class ProducerThread extends Thread{
    //private final Random random;
    private final Table table;
    private static int id=0;
    public ProducerThread(String name,Table table){
        super(name);
        this.table=table;
    }
    public void run(){
        try{
            while(true){
                Thread.sleep(1000);
                String cake="产品"+nextId();
                table.put(cake);
            }
        }catch(InterruptedException e){}
    }
    private static synchronized int nextId(){
        return id++;
    }
}
//消费者线程
class ConsumerThread extends Thread{
    private final Table table;
    public ConsumerThread(String name,Table table){
        super(name);
        this.table=table;
        
    }
    public void run(){
        try{
            while(true){
                String cake=table.take();
                Thread.sleep(2000);
            }
        }catch(InterruptedException e){}
    }
}
//缓冲池
class Table{
     private final String[] buffer;
     private int tail;
     private int head;
     private int count;
     public Table(int count){
         this.buffer=new String[count];
         this.head=0;
         this.tail=0;
         this.count=0;
     }
     synchronized void put(String cake)throws InterruptedException{
         System.out.println(Thread.currentThread().getName()+"  放入  "+cake);
         while(count>=buffer.length){
             System.out.println(Thread.currentThread().getName()+"  缓冲池满了，等一等吧...");
             wait();
             System.out.println(Thread.currentThread().getName()+"  有空间了，开始放...");
         }
         buffer[tail]=cake;
         tail=(tail+1)%buffer.length;
         count++;
         System.out.println("缓冲池中的产品数---->"+(count+1));
         notifyAll();
     }
     synchronized String take()throws InterruptedException{
         while(count<=0){
             System.out.println(Thread.currentThread().getName()+"  缓冲池没产品了，等一等吧...");
             wait();
             System.out.println(Thread.currentThread().getName()+"  有产品了，开吃...");
         }
         String cake=buffer[head];
         head=(head+1)%buffer.length;
         count--;
         notifyAll();
         System.out.println(Thread.currentThread().getName()+"  拿走  "+cake);
         System.out.println("缓冲池中的产品数----->"+(count+1));
         return cake;
     }
}
public class Main {
  public static void main(String args[]){
      Table table=new Table(10);
      new ProducerThread("生产者",table).start();
      new ConsumerThread("消费者",table).start();
  }
}
可以运行------>多多指正~
----------------解决方案--------------------------------------------------------
程序的优化是没有止境的,能运行出来就行,如果你这个只是交作业的话.如果是真实的系统,你按照客户的需求来好了。
----------------解决方案--------------------------------------------------------
这个比较简单

n个生产者

m个消费者

才更实际点

----------------解决方案--------------------------------------------------------
这个很简单啊：
new ProducerThread("生产者1",table).start();
new ProducerThread("生产者2",table).start();
new ProducerThread("生产者3",table).start();
new ProducerThread("生产者4",table).start();
new ProducerThread("生产者5",table).start();
new ConsumerThread("消费者1",table).start();
new ConsumerThread("消费者2",table).start();
new ConsumerThread("消费者3",table).start();
new ConsumerThread("消费者4",table).start();
不就行了


----------------解决方案--------------------------------------------------------