sleep和wait

验证:

1、在我们的第一个线程先start之后呢,我们的主线程就sleep 10 ms,次让第二个线程start.
* 那么第一个线程开始执行之后,获得同步锁,然后wait一秒钟,在第一个线程wait一秒钟的时候,第二个线程已经开始
* 执行,如果此时,我们的线程一并没有释放lock的话,那么线程二就会被阻塞,不能执行代码里面的逻辑,如果释放锁的话,
* 第二个线程是可以执行的。
public class WaitSleepDemo {
  public static void main(String[] args) {
    final Object lock = new Object();
    new Thread(new Runnable() {
      @Override
      public void run() {
        System.out.println("threada A waitting to a lock");
        synchronized (lock) {
          System.out.println("thread A get a lock");
          try {
            Thread.sleep(20);
            System.out.println("thread A wait method");
            lock.wait(1000);
            System.out.println("thread A is done");
          } catch (InterruptedException e) {
            e.printStackTrace();
          }
        }
      }
    }).start();

    //为了使第一段先执行
    try {
      Thread.sleep(10);
    } catch (InterruptedException e) {
      e.printStackTrace();
    }

    new Thread(new Runnable() {
      @Override
      public void run() {
        System.out.println("threada B waitting to a lock");
        synchronized (lock) {
          System.out.println("thread B get a lock");
          try {
            System.out.println("thread B sleep 10 ms");
            Thread.sleep(20);
            System.out.println("thread B is done");
          } catch (InterruptedException e) {
            e.printStackTrace();
          }
        }
      }
    }).start();
  }
}

  执行结果:

2、掉个个之后呢?可以看出B是连续执行的,也就是说B sleep一秒钟的同时,并没有放出资源,等到休眠玩一秒钟,又继续执行,B完全执行完才会释放出资源。

public class WaitSleepDemo {
  public static void main(String[] args) {
    final Object lock = new Object();
    new Thread(new Runnable() {
      @Override
      public void run() {
        System.out.println("threada B waitting to a lock");
        synchronized (lock) {
          System.out.println("thread B get a lock");
          try {
            System.out.println("thread B sleep 10 ms");
            Thread.sleep(1000);
            System.out.println("thread B is done");
          } catch (InterruptedException e) {
            e.printStackTrace();
          }
        }
      }
    }).start();
    //为了使第一段先执行
    try {
      Thread.sleep(10);
    } catch (InterruptedException e) {
      e.printStackTrace();
    }
    new Thread(new Runnable() {
      @Override
      public void run() {
        System.out.println("threada A waitting to a lock");
        synchronized (lock) {
          System.out.println("thread A get a lock");
          try {
            Thread.sleep(20);
            System.out.println("thread A wait method");
            lock.wait(1000);
            System.out.println("thread A is done");
          } catch (InterruptedException e) {
            e.printStackTrace();
          }
        }
      }
    }).start();
  }
}

  结果:

 

posted @ 2019-07-31 18:05 左手背右手背 阅读(...) 评论(...) 编辑 收藏