To generate a random int in the range [0, 1_000]: int n = new SplittableRandom().nextInt(0, 1_001); To generate a random int[100] array of values in the range [0, 1_000]: int[] a = new …

Using the Random class is the way to go as suggested in the accepted answer, but here is a less straight-forward correct way of doing it if you didn't want to create a new Random object : min …

Random r = new Random(); while (true) { // lower bound is 0 inclusive, upper bound is 49 exclusive // so we add 1 int n = r.nextInt(49) + 1; System.out.println("Number generated: "+n); …

May 5, 2011 · The first solution is to use the java.util.Random class: import java.util.Random; Random rand = new Random(); // Obtain a number between [0 - 49]. int n = rand.nextInt(50); // …

Jul 31, 2012 · First of, you have to create a Random object, such as: Random r = new Random(); And then, if you want an int value, you should use nextInt int myValue = r.nextInt(max); Now, if …

Random class has a method to generate random int in a given range. For example: Random r = new Random(); int x = r.nextInt(100); This would generate an int number more or equal to 0 …

Random rand = new Random(); // nextInt as provided by Random is exclusive of the top value so you need to add 1 int randomNum = rand.nextInt((max - min) + 1) + min; See the relevant …

In short, you can use the Random class which generates random integers, or you can use Math.Random and scale the answer then add your floor value. The second approach slightly …

Mar 11, 2011 · If You want to specify a more decent range, like from 10 to 100 ( both are in the range ) so the code would be : int Random =10 + (int)(Math.random()*(91)); /* int Random = …

3 days ago · (int) Math.floor(Math.random() * 101) Between one and hundred, I would do: (int) Math.ceil(Math.random() * 100) But what if I wanted to get a number between three and five? …