在Java中同步2个线程更简单的方法？

Question

我不确定在使用辅助线程执行某些代码之后,主线程中的某些代码将被执行.这是我得到的:

    final Object lock = new Object();
    final Thread t = new Thread(new Runnable() {
        public void run() {
            synchronized(lock) {
                System.out.println("qwerty");
                lock.notify();
            }
        }
    });

    synchronized(lock) {
        t.start();
        lock.wait();
    }

    System.out.println("absolutely sure, qwerty is above");

1. 这是正确的解决方案？
2. 有什么更短的方法吗？

Answer 1

事情是这样通知()和等待()是非常低级别的同步原语.

您应尽可能使用更高级别的抽象,例如,在CountDownLatch中.

以下只是一个可以帮助您入门的示例(例如,此处不考虑超时问题):

    final CountDownLatch latch = new CountDownLatch(1);
    final Thread t = new Thread(new Runnable() {
        public void run() {
            System.out.println("qwerty");
            latch.countDown();
        }
    });

    t.start();
    latch.await();

    System.out.println("absolutely sure, qwerty as been printed");

等待和通知等低级事物实际上只是您不应该关注的低级Java特性(除非您正在编写并发API).

另外,我建议阅读这本神奇的书:Java Concurrency In Practice.

Answer 2

假设您的主线程需要启动辅助线程的处理,Exchanger将是最简单的解决方案.

如果辅助线程是独立的,那么某种形式的BlockingQueue将是合适的.

使用Exchanger的示例(具有适当的异常处理).一个交换器可以用两个队列代替.

public static void main(String[] argv)
throws Exception
{
    final Exchanger<String> exchanger = new Exchanger<String>();
    new Thread(new Runnable() 
    {
        @Override
        public void run() 
        {
            try
            {
                String s = exchanger.exchange("");
                System.out.println(s);
                exchanger.exchange("this came from subthread");
            }
            catch (InterruptedException ex)
            {
                System.out.println("interrupted while waiting for message");
            }
        }
    }).start();

    exchanger.exchange("this came from main thread");
    String s = exchanger.exchange("");
    System.out.println(s);
}