Random Class

Yeah, I'm sorry. I was pretty tired when I read your code the first time. Looking at it now I understand it. It is VERY simple, so I guess I must have been VERY tired :) Thanks.

No, it wont. It will first return a random number between 0-5 and then look at your array at the position of the random number and return the number at that offset in your array. Best thing to do would be for you to make a small console application using this concept and run it under the debugger until you understand what's being done.

So, just to make sure: if I have an array with the following numbers (1,2,3,20,30,50) and use the code you provided, it will not return for example 4, 5, 10 or 45? Thanks guys!

use myArray.Length to get the ceiling, not count as in my code sample.

Ok. using System; Random random = new Random() // declare and fill your array ... int ceiling = myArray.Count; // How many elements in your array int arrayIndex = random.Next(ceiling); // Grab a random # between zero and ceiling int myRandom = myArray[arrayIndex]; // Whats the value in your array at the rand index?

Here you go: using System; class RandomEx { static void Main() { int[] numbers = {1,2,3,4,5}; Random r = new Random(); for(int i = 0; i < 20; i++) Console.WriteLine(numbers[r.Next(numbers.GetLowerBound(0), numbers.Length)]); } }