确定Java程序的确定性线程执行

我编写了一个非常简单的玩具程序，根据用户指定的时间要求完成线程内任务的执行。代码和示例输出如下。会发生什么情况是，每次运行代码时，任务完成时间都会在用户指定的时间的+ delta范围内。例如，如果用户想要在5秒内完成程序，则根据运行代码的CPU，它可以在5093或5012毫秒内完成。我想添加一些代码，它可以自动确定特定CPU为特定版本的JVM提供的最低延迟。根据该检测代码，可以将一个delta值添加到该行中，如：if ((taskRunTime > patience+delta) && t.isAlive())，以便系统为任务执行时间带来更高精度。请提出一些建议。确定Java程序的确定性线程执行

代码：

public class ThreadExample 
{ 


    static void threadMessage(String message) 
    { 
     String threadName = Thread.currentThread().getName(); 
     System.out.format("%s: %s%n", threadName, message); 
    } 

    private static class MessageLoop implements Runnable 
    { 
     public void run() 
     { 
      String importantInfo[] = 
      { 
       "A new integrated approach to programming", 
       "The innovative approach of the system", 
       "The input of a tracking system", 
       "A simulation system is then used for collision checking" 
      }; 
      try 
       { 
        for (int i = 0; i < importantInfo.length; i++) 
         { 

          Thread.sleep(4000); 
          threadMessage(importantInfo[i]); 
         } 
       } 
       catch (InterruptedException e) 
        { 
         threadMessage("I wasn't done!"); 
        } 
     } 
    } 

    public static void main(String args[]) throws InterruptedException 
    { 


     //Delay, in milliseconds before we interrupt MessageLoop 
     long patience = 1000 * 60 * 60; 

     //If command line argument present, gives patience in seconds. 
     if (args.length > 0) 
     { 
      try { 
       patience = Long.parseLong(args[0]) * 1000; 
      } catch (NumberFormatException e) { 
       System.err.println("Argument must be an integer."); 
       System.exit(1); 
      } 

     } 

     threadMessage("Starting MessageLoop thread"); 
     long startTime = System.currentTimeMillis(),taskRunTime=0; 
     Thread t = new Thread(new MessageLoop()); 
     t.start(); 

     threadMessage("Waiting for MessageLoop thread to finish"); 
     //loop until MessageLoop thread exits 
     while (t.isAlive()) 
     { 
      threadMessage("Still waiting..."); 
      //Wait maximum of 1 second for MessageLoop thread to finish. 
      t.join(100); 
      taskRunTime=System.currentTimeMillis() - startTime; 
      if ((taskRunTime > patience) && t.isAlive()) 
      { 
       threadMessage("Tired of waiting...task is running longer than the patience you set or the default!"); 
       t.interrupt(); 
       t.join(); 
      } 

     } 
     threadMessage("Finally out of thread!"); 
     System.out.println("Time to complete task="+taskRunTime+"ms"); 

    } 
}

从英特尔迅驰1.7 Ghz的计算机输出样品（Java的热点（TM）客户端虚拟机（建设10.0 - B23，混合模式））

java -jar ThreadExample.jar 5 
main: Starting MessageLoop thread 
main: Waiting for MessageLoop thread to finish 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
Thread-0: A new integrated approach to programming 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Still waiting... 
main: Tired of waiting...task is running longer than the patience you set or the default! 
Thread-0: I wasn't done! 
main: Finally out of thread!

来源

2009-08-11 iceman