Question

是的，这是一项家庭作业，但我已经尽了一切可能，但却找不到可能。这个任务的重点是说明在实现dekker算法/ peterson算法之前，两个进程很可能不会一个接一个地进行。

 import java.util.*;

public class myProcess
{

     private static final Random R = new Random();
     private int id;


     public myProcess(int i){
         id = i;
    }


    private static void delay(int value){

         try{
             java.lang.Thread.sleep(R.nextInt(value));
         }
         catch(InterruptedException e){
         }
     }

     public void run(){              

         System.out.println("");
         delay(20);
         System.out.println(this.id + " is starting");
         delay(20);
         System.out.println("LINE ONE");
         delay(20);
         System.out.println("LINE TWO");
         delay(20);
         System.out.println("LINE THREE");
         delay(20);
         System.out.println(this.id+ " is ending ");
         delay(20);

     }

     public static void main(String [] args){
         final int N = 2;
         myProcess[] t = new myProcess[N];

         for(int i = 0; i < N; i++){
             t[i] = new myProcess(i);
             t[i].run();

         }


  }

现在输出

 0 is starting
 LINE ONE
 LINE TWO
 LINE THREE
 0 is ending
 1 is starting
 LINE ONE
 LINE TWO
 LINE THREE
 1 is ending

但它应该是混合起来，以说明进程不一定等待另一个进程完成。

我尝试了其他定义run（）的方法，例如

 String[] statements = new String[5];
         statements[0] = "Thread " + this.id + " is starting iteration ";
         statements[1] = "We hold these truths to be self-evident, that all men are          created equal,";
         statements[2] = "that they are endowed by their Creator with certain unalienable Rights,";
         statements[3] = "that among these are Life, Liberty and the pursuit of Happiness.";
         statements[4] = "Thread " + this.id+ " is done with iteration ";

    for(int i = 0; i< 5; i++){

        System.out.println(statements[i]);
        delay(20);
    }

但它仍然没有给我任何“错误的输出”

我做错了什么让输出如此正确？

Answer 1

您应该在线程上调用start()函数，而不是run()。

编辑：您的类也应该实现Runnable接口或扩展Thread类。

您没有在代码中创建新的线程，并且所有内容都在一个线程中运行。

public class myProcess extends Thread

...

for(int i = 0; i < N; i++){
    t[i] = new myProcess(i);
    t[i].start();
}

Answer 2

我猜你的延迟时间太短，看不到任何明显的混音。你传入20，好像它是20秒，但它只有20毫秒的睡眠。通过20,000，看看你是否得到了你期望的行为。

Answer 3

将延迟方法更改为如下所示。根据这篇文章（https://stackoverflow.com/a/1600603/1265692），Java sleep方法不能保证放弃对cpu的控制。通过添加yield调用，您可以提醒Java让其他线程运行。

private static void delay(int value){

     try{
         java.lang.Thread.sleep(R.nextInt(value));
         java.lang.Thread.yield();
     }
     catch(InterruptedException e){
     }
 }

即使没有dekker / pterson算法，Java中的多线程也可以工作？

3 个答案: