How do you evaluate getTime() repeatedly?

[Art Metzer]

Hello, everyone.
At http://www.norvig.com/java-iaq.html#math, Peter Norvig claims that counting down (starting at n and decrementing until 0 in a for loop, for example) executes twice as fast as counting up (starting at 0 and incrementing until n). I believe him, but I wanted to see this bit of truth play out for myself.
I then wrote and tried this bit of code:

import java.util.*;
public class TimeCheck {
    public static void main( String[] args ) {
        long now = ( new Date() ).getTime();
        long stop = now + 1000;  // One second in the future
        long ell;                // Number of increments
             
        for( ell = 0; ( new Date() ).getTime() < stop; ell++ ) {
        }
                            
        System.out.println( ell );
    }
}

I got results in the thousands (instead of the millions), and I figure that's because every iteration of my loop, Java incurs the overhead of having to instantiate java.util.Date.
Is there a more efficient way to poll the time every iteration of my for loop? Or, better yet, can you produce source code that would more efficiently/clearly demonstrate Mr. Norvig's statement?
Thank you, one and all.
Art
a.k.a. Curious George

[Grant Crofton]

You can use System.currentTimeMillis() to get a long (I think it's the number of MilliSeconds since 1980, or something).
I tried this:

public class TimeCheck
{
   public static void main( String args[] )
   {
      long startTime = System.currentTimeMillis();
      long endTime = 0;
      
      for( int count=0; count<100000000; count++ )
      {}
      
      endTime = System.currentTimeMillis();
      
      System.out.println( "Counting up took " + (endTime - startTime) + " MS." );
      
      startTime = System.currentTimeMillis();
      
      for( int count=100000000; count>0; count-- )
      {}
      
      endTime = System.currentTimeMillis();
      
      System.out.println( "Counting down took " + (endTime - startTime) + " MS." );
   }
}

You'd have to run it a few times to get a fair result. I'll let you see what the answer is.