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Thread: Random number help
 09152008, 05:28 PM #1Member
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Random number help
Hi all,
I need help with a random number feature. I want a random number between 1 and 26 (including 1 and 26). I have this
letter =(int) (Math.random() * 26 + 1);
Since Math.random() is 0  .9999999, how do i get whole numbers?
should I do this:
so right now i am not getting accurate whole numbers. can anyone help?
Thanks
 09152008, 05:33 PM #2
Random has a nextInt() that takes an upper bound. There are short and well known ways to get the number in a range.
Introduction to Programming Using Java.
Cybercartography: A new theoretical construct proposed by D.R. Fraser Taylor
 09152008, 05:36 PM #3Member
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I was looking into that but I have to keep the Random() function. Thanks tho. Is it possible to get some insight with the Random feature?
 09152008, 06:34 PM #4
basically simple
Well, in general what we do is int num = Math.random() times range plus floor. It is simple, you just have to do a timeout and think about it.
Some number times ( 0  1 ) gives, bascially, zero to the number. Then we often have a range so it is upper limit minus lower limit times random plus floor.Introduction to Programming Using Java.
Cybercartography: A new theoretical construct proposed by D.R. Fraser Taylor
 09162008, 04:13 AM #5Member
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I created a quick test class and only got numbers between 1 and 26 inclusive using basically what you posted.
Java Code:public class TestRandom { /** * @param args */ public static void main(String[] args) { for (int i = 0; i < 50000; i++) { int letter = (int)(Math.random() * 26 + 1); System.out.println(letter); if(letter == 0  letter > 26){ throw new RuntimeException("The value is out of range" + letter); } } } }
 09162008, 09:13 AM #6
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You can do this easily in this way as well.
Java Code:public static void main(String[] args) { Random randomGenerator = new Random(); for(int i = 0; i < 15; i++) { int j = randomGenerator.nextInt(26); System.out.println(j + 1); } }
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