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由Thread.join引发的思考

时间:2020-02-22 16:06:02      阅读:71      评论:0      收藏:0      [点我收藏+]

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下面是一段司空见惯的代码,创建两个线程A和线程B,使得线程A优先于线程B执行,使得线程B优先于主线程执行

public class Demo52 {
    public static void main(String[] args) {
         
        Thread thread1 = new Thread(()->{
            System.out.println("线程:"+Thread.currentThread().getName());
        },"A");
        Thread thread2 = new Thread(()->{
            try {
                thread1.join();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            System.out.println("线程:"+Thread.currentThread().getName());
        },"B");
        thread1.start();
        thread2.start();
        try {
            thread2.join();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("main线程");
    }
}

输出结果:

线程:A
线程:B
main线程

它是如何做到的线程A优先于线程B,线程B优先于主线程的呢?

为了说明这点,就要查看Thread.join的源码了:

     /** 等待该线程终止
     * Waits for this thread to die.
     * 调用此方法的行为方式与调用完全相同join (0)
     * <p> An invocation of this method behaves in exactly the same
     * way as the invocation
     *
     * <blockquote>
     * {@linkplain #join(long) join}{@code (0)}
     * </blockquote>
     *
     * @throws  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 void join() throws InterruptedException {
        join(0);
}
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;
            }
        }
}
public final native void wait(long timeout) throws InterruptedException;

? 结合上面的实例,先看主线程和线程B,在主线程中执行“thread2.join();”,也就相当进入到join方法体中,它会获取到当前实例的锁,也就是线程B对象的锁,然后判断线程是否存活后执行“ wait(0);”,执行该方法时,主线程会释放锁,进入到阻塞状态(也即进入到了该锁的WaitSet中)。为什么说是主线程进入到阻塞状态,而不是线程B进入到阻塞状态呢?

为了解答这个问题,设计如下的实例:

class ThreadTest extends Thread{
    public synchronized void method1(){
        System.out.println("hello world");
        System.out.println(Thread.currentThread().getName());
    }
}
public class Demo51 {
    public static void main(String[] args) {
        ThreadTest threadTest = new ThreadTest();
        threadTest.start();
        threadTest.method1();
    }
}

输出结果:

hello world
main

? 可以看到在执行“threadTest.method1()”时,线程“threadTest”中输出的当前线程是主线程,而不是“Thread-0”,这是因为它是“ threadTest.method1()”是由主线程所调用的。

? 再回到上面的问题中,调用“thread2.join();”导致主线程被阻塞,紧接着线程B开始执行,线程B执行完毕后开始执行主线程,而此时主线程还在阻塞状态(即还在该线程锁的waitset中),那么它是如何实现唤醒主线程,使得它能够接着执行的呢?实际上这是因为每个线程在退出时,会执行notifyAll唤醒所有阻塞在该实例锁上的线程。

为了更为方便的说明这个问题,设计如下的实例,创建一个线程类,定义“method1”方法和“run”方法,然后在主线程中调用它

class ThreadTest extends Thread{

    public synchronized void method1() throws InterruptedException {
        System.out.println("hello world");
        System.out.println(Thread.currentThread().getName());
        wait();
        System.out.println("==================");
    }

    @Override
    public void run() {
        try {
            Thread.sleep(10_000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }
}
public class Demo51 {
    public static void main(String[] args) throws InterruptedException {
        ThreadTest threadTest = new ThreadTest();
        threadTest.start();
        threadTest.method1();
        System.out.println("===Exit===");
    }
}

运行结果:

hello world
main
==================
===Exit===

? 来分析一下程序的执行,“ThreadTest threadTest = new ThreadTest()”创建线程对象,“threadTest.start();”启动这个线程,“threadTest.method1();”执行method1方法,关注点就在这里,在执行method1方法的时候,主线程会尝试获取“threadTest ”对象的锁,成功后进入到方法体,然后进入到阻塞状态,而threadTest 在启动后执行run方法中的内容,然后睡眠10秒钟,在随眠结束后执行完毕,threadTest 退出执行状态,在退出时,执行notify_all方法唤醒阻塞在threadTest 锁上的主线程。

// 位于/hotspot/src/share/vm/runtime/thread.cpp中

void JavaThread::exit(bool destroy_vm, ExitType exit_type) {
// ...
// Notify waiters on thread object. This has to be done after exit() is called
// on the thread (if the thread is the last thread in a daemon ThreadGroup the
// group should have the destroyed bit set before waiters are notified).
// 有一个贼不起眼的一行代码,就是这行
ensure_join(this);
// ...
}
static void ensure_join(JavaThread* thread) {
// We do not need to grap the Threads_lock, since we are operating on ourself.
Handle threadObj(thread, thread->threadObj());
assert(threadObj.not_null(), "java thread object must exist");
ObjectLocker lock(threadObj, thread);
// Ignore pending exception (ThreadDeath), since we are exiting anyway
thread->clear_pending_exception();
// Thread is exiting. So set thread_status field in  java.lang.Thread class to TERMINATED.
java_lang_Thread::set_thread_status(threadObj(), java_lang_Thread::TERMINATED);
// Clear the native thread instance - this makes isAlive return false and allows the join()
// to complete once we've done the notify_all below
java_lang_Thread::set_thread(threadObj(), NULL);

// 同志们看到了没,别的不用看,就看这一句
// thread就是当前线程,是啥?就是刚才例子中说的threadA线程啊。
lock.notify_all(thread);

// Ignore pending exception (ThreadDeath), since we are exiting anyway
thread->clear_pending_exception();
}

————————————————
版权声明:本文为CSDN博主「Mlib」的原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接及本声明。
原文链接:https://blog.csdn.net/u010983881/article/details/80257703

? 同理在Thread.join中,也是如此。如果一个线程A执行了thread.join()语句,其含义是:当前线程A等待thread线程终止之后才从thread.join()返回;它的底层实现就是线程A进入到了thread对象的waitset中了,当thread的线程执行完毕后,在线程退出操作中,会自动唤醒阻塞在thread对象上的线程A,这样线程A也就能够继续执行了,表现为线程A等待thread执行完毕后,才接着执行。

参考链接:Java Thread的join() 原理

【Java】Thread类中的join()方法原理

(四)Thread.join的作用和原理

由Thread.join引发的思考

标签:share   star   java   sam   状态   delay   run方法   是否存活   tin   

原文地址:https://www.cnblogs.com/cosmos-wong/p/12345299.html

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