Solved

UNIQUE random number

Posted on 2004-10-03
11
231 Views
Last Modified: 2010-03-31
I have a constructor that needs to assign a unique random number to one of its variable everytime it is been constructed. Can someone tell me how can that be done? I mean, it has to be unique. hm~~~ Does normal random number generator guarantee unique number??
           
0
Comment
Question by:jtcy
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 3
  • 3
  • 3
  • +1
11 Comments
 
LVL 92

Expert Comment

by:objects
ID: 12210652
no a random number generator will not guarantee uniquness.
you could store all the generated numbers in a Set, and check if it already exists using that (generating a new one if it did.
0
 

Author Comment

by:jtcy
ID: 12210670
so smth like...


------------------------------------


import java.util.Random;

public class Key
  {
     private int randomNumber;
     Set mySet = new Set();
   
     public Key()
      {
          Random generator = new Random();
          randomNumber = generator.nextInt();
          while (mySet.contains(randomNumber))
                randomNumber = generator.nextInt();
      }
}
               

-------------------------------------------------------------------
?

0
 
LVL 92

Expert Comment

by:objects
ID: 12210682
yep
0
VIDEO: THE CONCERTO CLOUD FOR HEALTHCARE

Modern healthcare requires a modern cloud. View this brief video to understand how the Concerto Cloud for Healthcare can help your organization.

 

Author Comment

by:jtcy
ID: 12210687
oops. Should be:

------------------------------------


import java.util.Random;

public class Key
  {
     private int randomNumber;
     Set mySet = new Set();
   
     public Key()
      {
          Random generator = new Random();
          randomNumber = generator.nextInt();
          while (mySet.contains(randomNumber))
                randomNumber = generator.nextInt();
          mySet.add(randomNumber);
      }
}
               

-------------------------------------------------------------------
0
 
LVL 92

Expert Comment

by:objects
ID: 12210689
you need to use the Integer wrapper class so you can add them to the set.
0
 

Author Comment

by:jtcy
ID: 12210694
Do u mean...

mySet.add((Integer)randomNumber);


0
 
LVL 86

Expert Comment

by:CEHJ
ID: 12210776
mySet.add(new Integer(randomNumber));
0
 
LVL 21

Accepted Solution

by:
MogalManic earned 20 total points
ID: 12211163
I all you want is a number with a high probability that it is unique, do the following:

   private static Random generator=new Random();
   public int getKey()
   {
          return generator.nextInt() ^ System.currentTimeMillis()
   }
  The current time in millis will be unique as long as two threads don't call getKey at the same time.  If so, the random number should ensure uniqueness.
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 12211183
That would have to be:


return generator.nextInt() ^ (int)System.currentTimeMillis();
0
 
LVL 21

Expert Comment

by:MogalManic
ID: 12211314
no, Javas automatic conversion would be more correct  the following is what java does automatically:
    return (int) ((long) generator.nextInt() ^ System.currentTimeMillis());

OR if you did not want to drop bits
private static Random generator=new Random();
   public long getKey()
   {
         return generator.nextLong() ^ System.currentTimeMillis()
   }
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 12211415
>>the following is what java does automatically

Certainly doesn't in the version of sdk i'm using (1.5.0-beta2-b51)
0

Featured Post

Optimize your web performance

What's in the eBook?
- Full list of reasons for poor performance
- Ultimate measures to speed things up
- Primary web monitoring types
- KPIs you should be monitoring in order to increase your ROI

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

INTRODUCTION Working with files is a moderately common task in Java.  For most projects hard coding the file names, using parameters in configuration files, or using command-line arguments is sufficient.   However, when your application has vi…
For beginner Java programmers or at least those new to the Eclipse IDE, the following tutorial will show some (four) ways in which you can import your Java projects to your Eclipse workbench. Introduction While learning Java can be done with…
Viewers will learn about if statements in Java and their use The if statement: The condition required to create an if statement: Variations of if statements: An example using if statements:
This tutorial explains how to use the VisualVM tool for the Java platform application. This video goes into detail on the Threads, Sampler, and Profiler tabs.
Suggested Courses

626 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question