sleep()方法如何在给定的线程和输出上工作？

Question

据我所知,sleep()它用于在指定时间内休眠一个线程.我制成的纤维束的例子-比如1我正在作为输出1,2,3,4因为我只创建一个.在例2中,我创建的线程的2个实例,我获得输出1,1,2,2,3,3,4,4.

为什么输出不1,2,3,4第一个线程,然后1,2,3,4为第二个？

例1:

// Using sleep() method
  public class Aaa extends Thread{
    public void run(){
        for(int i=1;i<5;i++){
        try{
           Thread.sleep(500); 
        } catch(InterruptedException e){
            System.out.println(e);
        }
        System.out.println(i);
        }
    }
    public static void main(String[] args){
        Aaa m1=new Aaa();
        m1.start();
    }
  }

Output:
    1
    2
    3
    4

例2:

    // Using sleep() method
  public class Aaa extends Thread{
    public void run(){
        for(int i=1;i<5;i++){
        try{
           Thread.sleep(500); // sleeps thread
        } catch(InterruptedException e){
            System.out.println(e);
        }
        System.out.println(i);
        }
    }
    public static void main(String[] args){        
       Aaa m1=new Aaa(); // creating one object
       Aaa m2=new Aaa(); // creating second object of a class
       m1.start(); // calls run method
       m2.start();
    }
  }


Output:
    1
    1
    2
    2
    3
    3
    4
    4

Answer 1

您已创建了两个Runnable对象.如果你通过调用他们的run方法运行它们,你会得到你想象的:

   Aaa m1=new Aaa(); // creating one object
   Aaa m2=new Aaa(); // creating second object of a class
   System.out.println("Calling m1.run()");
   m1.run(); // we call run method
   System.out.println("Calling m2.run()");
   m2.run();

产量

Calling m1.run()
1
2
3
4
Calling m2.run()
1
2
3
4

因为该方法执行两次,一个接一个.

需要注意的是调用run的方法Runnable/Thread是不运行一个线程的正常方式.您通常会使用该start方法.

但是,如果你把每个人放在一个Thread和start它们中:

   Aaa m1=new Aaa(); // creating one object
   Aaa m2=new Aaa(); // creating second object of a class
   System.out.println("Calling m1.start()");
   m1.start(); // thread calls run method
   System.out.println("Calling m2.start()");
   m2.start();

现在,每个对象并行运行在自己的线程中,因此输出是交错的:

Calling m1.start()
Calling m2.start()
1 < From thread 1
1 < From thread 2
2 ...
2
3
3
4
4

顺序显然可以根据线程如何交错而变化.

可能令你困惑的一件事是你选择了extend Thread.这是不鼓励的.最好implement Runnable- 像这样:

public class Aaa implements Runnable {
  public void run() {
    for (int i = 1; i < 5; i++) {
      try {
        Thread.sleep(500); // sleeps thread
      } catch (InterruptedException e) {
        System.out.println(e);
      }
      System.out.println(i);
    }
  }

  public static void main(String[] args) {
    Aaa m1 = new Aaa(); // creating one object
    Thread t1 = new Thread(m1); // Its thread
    Aaa m2 = new Aaa(); // creating second object of a class
    Thread t2 = new Thread(m2); // Its thread
    t1.start(); // calls m's run method in a new thread.
    t2.start();
  }

}

现在更清楚的是,您的对象在两个不同的线程中运行,因此并行运行.