多线程编程之join()方法

现实生活中,有些工作是需要团队中成员依次完成的,这就涉及到了一个顺序问题。现在有T1、T2、T3三个工人,如何保证T2在T1执行完后执行,T3在T2执行完后执行?

问题分析:首先问题中有三个实体,T1、T2、T3, 因为是多线程编程,所以都要设计成线程类。关键是怎么保证线程能依次执行完呢?

 

Java实现过程如下:

public class T1 implements Runnable{

	@Override
	public void run() {
	    try {
			System.out.println("T1开始工作.....");
			Thread.sleep(RandomUtils.nextInt(300));
			System.out.println("T1结束工作>>>>>");
		} catch (InterruptedException e) {
			e.printStackTrace();
		}
	}

}

 

package thread.join.demo1;

import org.apache.commons.lang.math.RandomUtils;

public class T2 implements Runnable{

	@Override
	public void run() {
		try{
			System.out.println("T2开始工作.....");
			Thread.sleep(RandomUtils.nextInt(300));
			System.out.println("T2结束工作>>>>>");
		} catch (InterruptedException e) {
			e.printStackTrace();
		}
	}

}

 

public class T3 implements Runnable{

	@Override
	public void run() {
		try{
			System.out.println("T3开始工作.....");
			Thread.sleep(RandomUtils.nextInt(300));
			System.out.println("T3结束工作>>>>>");
		} catch (InterruptedException e) {
			e.printStackTrace();
		}
	}

}

 

public class Main {

	public static void main(String[] args){
		
		Thread t1 = new Thread(new T1());
		Thread t2 = new Thread(new T2());
		Thread t3 = new Thread(new T3());
		
		t1.start();
		t2.start();
		t3.start();
		
		System.out.println("T1、T2、T3依次工作结束.");
	}

}
 

运行结果:

T1开始工作.....

T2开始工作.....

T3开始工作.....

T1T2T3依次工作结束.

T3结束工作>>>>>

T2结束工作>>>>>

T1结束工作>>>>>

 

说明:从结果来看,T1T2T3并没有依次执行。查看JDK文档,java.lang.Thread 类有三个join()方法,其解释为:等待该线程终止。试用它来解决该问题……

 

Main.java修改如下:

public class Main {

	public static void main(String[] args) throws InterruptedException {
		
		Thread t1 = new Thread(new T1());
		Thread t2 = new Thread(new T2());
		Thread t3 = new Thread(new T3());
		
		t1.start();
		t1.join();
		
		t2.start();
		t2.join();
		
		t3.start();
		t3.join();
		
		System.out.println("T1、T2、T3依次工作结束.");
	}

}

 

 

运行结果:

T1开始工作.....

T1结束工作>>>>>

T2开始工作.....

T2结束工作>>>>>

T3开始工作.....

T3结束工作>>>>>

T1T2T3依次工作结束.

 

 

哈,解决了!

 

 

查看jdk源码,其中join方法代码片断如下:

 

/**
     * Waits at most <code>millis</code> milliseconds for this thread to 
     * die. A timeout of <code>0</code> means to wait forever. 
     *
     * @param      millis   the time to wait in milliseconds.
     * @exception  InterruptedException if any thread has interrupted
     *             the current thread.  The <i>interrupted status</i> of the
     *             current thread is cleared when this exception is thrown.
     */
    public final synchronized void join(long millis) 
    throws InterruptedException {
	long base = System.currentTimeMillis();
	long now = 0;

	if (millis < 0) {
            throw new IllegalArgumentException("timeout value is negative");
	}

	if (millis == 0) {
	    while (isAlive()) {
		wait(0);
	    }
	} else {
	    while (isAlive()) {
		long delay = millis - now;
		if (delay <= 0) {
		    break;
		}
		wait(delay);
		now = System.currentTimeMillis() - base;
	    }
	}
    } 

 

 

单纯从代码上看如果线程被生成了,但还未被起动,isAlive()将返回false,调用它的join()方法是没有作用的,将直接继续向下执行。Main.java类中,t1.join()是判断t1线程的状态,如果t1isActive()方法返回false,在 t1.join(),这一点就不用阻塞了,可以继续向下进行了。从源码里看,wait方法中有参数,也就是不用唤醒谁,只是不再执行wait,向下继续执行而已。join()方法中,对于isAlive()wait()方法的作用对象是个比较让人困惑的问题:

isAlive()方法的签名是:public final native boolean isAlive(),也就是说isAlive()是判断当前线程的状态,也就是t1的状态。

 

 

 

 

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转载自15838341661-139-com.iteye.com/blog/2225008