Getting random numbers in Java - Stack Overflow
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); // Add 1 to the result to get a number from the required range. // (i.e., [1 - 50]). n += 1; Another solution is using Math.random():
How do I generate random integers within a specific range in Java ...
Java 7+. In Java 1.7 or later, the standard way to do this (generate a basic non-cryptographically secure random integer in the range [min, max]) is as follows: import java.util.concurrent.ThreadLocalRandom; // nextInt is normally exclusive of the top value, // so add 1 to make it inclusive. int randomNum = ThreadLocalRandom.current().nextInt ...
Java random numbers using a seed - Stack Overflow
Sep 17, 2012 · This is my code to generate random numbers using a seed as an argument: double randomGenerator(long seed) { Random generator = new Random(seed); double num = generator.nextDouble() * (0.5)...
Java Generate Random Number Between Two Given Values
Mar 11, 2011 · Java doesn't have a Random generator between two values in the same way that Python does. It actually only takes one value in to generate the Random. What you need to do, then, is add ONE CERTAIN NUMBER to the number generated, which will cause the number to be within a range.
Is there functionality to generate a random character in Java?
Apr 13, 2010 · java.util.Random is the more effective one I have tried out so far, having a precision of 98.65% uniqueness. I have provided bellow some tests which generate 10000 batches of a hundred 2 alphanumeric chars strings and calculates the average. Other random tools were RandomStringUtils from commons.lang3 and java.util.Math.
Java random number with given length - Stack Overflow
Mar 22, 2011 · To generate a 6-digit number: Use Random and nextInt as follows: Random rnd = new Random(); int n = 100000 + rnd.nextInt(900000); Note that n will never be 7 digits (1000000) since nextInt(900000) can at most return 899999.
java - How to randomly pick an element from an array - Stack …
15. If you are going to be getting a random element multiple times, you want to make sure your random number generator is initialized only once. private int[] items = new int[]{1,2,3}; private Random rand = new Random(); public int getRandArrayElement(){. return items[rand.nextInt(items.length)]; If you are picking random array elements that ...
How to generate a random String in Java - Stack Overflow
May 19, 2010 · Generating a random string of characters is easy - just use java.util.Random and a string containing all the characters you want to be available, e.g. public static String generateString(Random rng, String characters, int length) {. char[] text = new char[length]; for (int i = 0; i < length; i++)
How Java random generator works? - Stack Overflow
Feb 17, 2016 · The Random class is implemented as a linear congruential number generator (LCNG). The general formula for a linear congruential generator is: new_state = (old_state * C1 + C2) modulo N. The precise algorithm used by Random is specified in the javadocs. If you know the current state of the generator 1, the next state is completely predictable.
java - How to create an array of 20 random bytes? - Stack Overflow
Apr 15, 2011 · 4. Create a Random object with a seed and get the array random by doing: public static final int ARRAY_LENGTH = 20; byte[] byteArray = new byte[ARRAY_LENGTH]; new Random(System.currentTimeMillis()).nextBytes(byteArray); // get fisrt element.