That’s because a long is 64 bits and an int is 32 bits, not to mention you’re going from floating-point to integer. To go from long to int, you’re going to have to discard some information, and the compiler can’t/won’t do that automatically. You’re going to have to explicitly say so through a cast: int g = (int) Math.round(Math.random() * 255); ^^^^^ … Read more
You could use e.g. r.nextInt(101) For a more generic “in between two numbers” use: Random r = new Random(); int low = 10; int high = 100; int result = r.nextInt(high-low) + low; This gives you a random number in between 10 (inclusive) and 100 (exclusive)
Using Collections to shuffle an array of primitive types is a bit of an overkill… It is simple enough to implement the function yourself, using for example the Fisher–Yates shuffle: import java.util.*; import java.util.concurrent.ThreadLocalRandom; class Test { public static void main(String args[]) { int[] solutionArray = { 1, 2, 3, 4, 5, 6, 16, 15, 14, … Read more
Here you can use my method for generating Random String protected String getSaltString() { String SALTCHARS = “ABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890”; StringBuilder salt = new StringBuilder(); Random rnd = new Random(); while (salt.length() < 18) { // length of the random string. int index = (int) (rnd.nextFloat() * SALTCHARS.length()); salt.append(SALTCHARS.charAt(index)); } String saltStr = salt.toString(); return saltStr; } … Read more
The simplest way would be to create a list of the possible numbers (1..20 or whatever) and then shuffle them with Collections.shuffle. Then just take however many elements you want. This is great if your range is equal to the number of elements you need in the end (e.g. for shuffling a deck of cards). … Read more
Generate a random double in a range
You could use boolean values of 0 or 1 based on value of Math.random() as a double between 0.0 and 1.0 and make the random generator much simpler. And you can get rid completely of the coinToss() method. if(Math.random() < 0.5) { answer = “Tails”; tails++; } Remove the coin toss method and replace the first conditional with the … Read more
