java 多线程 生产者和消费者有关问题

java 多线程 生产者和消费者问题

 

用一个数据来表示数据存储空间

 

package com.yan.test;

public class SyncStack {

	private  int index = 0;

	private char[] stack = new char[6];

	public synchronized void push(char temp) {

		if (index == stack.length) {
			try {
				this.wait();
			} catch (InterruptedException e) {
				e.printStackTrace();
			}
		}
		this.notify();
		stack[index] = temp;
		index++;

		System.out.println(Thread.currentThread().getName() + " push " + temp);

	}

	public synchronized char pop() {

		if (index == 0) {
			try {
				this.wait();
			} catch (InterruptedException e) {
				e.printStackTrace();
			}
		}
		this.notify();

		index--;
		System.out.println("___" + Thread.currentThread().getName() + " ---pop " + stack[index]);
		return stack[index];

	}
}

 

 

生产者

 

 

package com.yan.test;

public class Produce implements Runnable {

	private SyncStack stack;

	int i = 0;

	public Produce(SyncStack stack) {
		super();
		this.stack = stack;
	}

	public void run() {

		while (true) {

			synchronized (this) {

				if (i < 28) {
					stack.push((char) ((char) i + 'A'));
					i++;
					
				} else {
					break;
				}
				
			}

		}

	}
}

 

消费者

 

 

 

package com.yan.test;

public class Consumer implements Runnable {

	private SyncStack stack;

	int i = 0;

	public Consumer(SyncStack stack) {
		super();
		this.stack = stack;
	}

	public void run() {

		while (true) {

			synchronized (this) {
				if (i < 28) {
					stack.pop();
					i++;
				} else {
					break;
				}
			}

		}

	}

}

 

 

 

测试的main方法

 

package com.yan.test;

public class TestSysn {

	public static void main(String[] args) {
		SyncStack s = new SyncStack();
		Produce p = new Produce(s);

		Consumer c = new Consumer(s);
		Thread t1 = new Thread(p);
		t1.setName("product1");
		Thread t3 = new Thread(p);
		t3.setName("product2");
		Thread t5 = new Thread(p);
		t5.setName("product3");
		t3.start();
		t1.start();
		t5.start();
		
		
		Thread t2 = new Thread(c);
		t2.setName("Consumer1");
		Thread t4 = new Thread(c);
		t4.setName("Consumer2");
	
		t2.start();
		t4.start();

	}

}

 

结果

product2 push A
product1 push B
product2 push C
product1 push D
product2 push E
product1 push F
___Consumer1 ---pop F
product2 push G
___Consumer1 ---pop G
___Consumer1 ---pop E
___Consumer1 ---pop D
___Consumer1 ---pop C
___Consumer1 ---pop B
___Consumer1 ---pop A
product1 push H
___Consumer1 ---pop H
product3 push I
___Consumer1 ---pop I
product2 push J
___Consumer1 ---pop J
product1 push K
___Consumer1 ---pop K
product3 push L
___Consumer1 ---pop L
product2 push M
___Consumer1 ---pop M
product1 push N
___Consumer2 ---pop N
product3 push O
___Consumer1 ---pop O
product2 push P
___Consumer2 ---pop P
product1 push Q
___Consumer1 ---pop Q
product3 push R
___Consumer2 ---pop R
product2 push S
___Consumer1 ---pop S
product1 push T
___Consumer2 ---pop T
product3 push U
___Consumer1 ---pop U
product2 push V
___Consumer2 ---pop V
product1 push W
___Consumer1 ---pop W
product3 push X
___Consumer2 ---pop X
product2 push Y
___Consumer1 ---pop Y
product1 push Z
product3 push [
___Consumer2 ---pop [
product2 push \
___Consumer1 ---pop \
___Consumer2 ---pop Z