Java: random long number in 0 <= x < n range

Question

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 and less than 100. I'd like to do exactly the same with long number.

long y = magicRandomLongGenerator(100);

Random class has only nextLong(), but it doesn't allow to set range.

Answer 1

ThreadLocalRandom

ThreadLocalRandom has a nextLong(long bound) method.

long v = ThreadLocalRandom.current().nextLong(100);

It also has nextLong(long origin, long bound) if you need an origin other than 0. Pass the origin (inclusive) and the bound (exclusive).

long v = ThreadLocalRandom.current().nextLong(10,100); // For 2-digit integers, 10-99 inclusive.

SplittableRandom has the same nextLong methods and allows you to choose a seed if you want a reproducible sequence of numbers.

Answer 2

How about this:

public static long nextLong(@NonNull Random r, long min, long max) {
    if (min > max)
        throw new IllegalArgumentException("min>max");
    if (min == max)
        return min;
    long n = r.nextLong();
    //abs (use instead of Math.abs, which might return min value) :
    n = n == Long.MIN_VALUE ? 0 : n < 0 ? -n : n;
    //limit to range:
    n = n % (max - min);
    return min + n;
}

?

Answer 3

The below Method will Return you a value between 10000000000 to 9999999999

long min = 1000000000L
long max = 9999999999L    

public static long getRandomNumber(long min, long max){

    Random random = new Random();         
    return random.nextLong() % (max - min) + max;

}

Answer 4

The methods above work great. If you're using apache commons (org.apache.commons.math.random) check out RandomData. It has a method: nextLong(long lower, long upper)

http://commons.apache.org/math/userguide/random.html

http://commons.apache.org/math/api-1.1/org/apache/commons/math/random/RandomData.html#nextLong(long,%20long)

Answer 5

Use the '%' operator

resultingNumber = (r.nextLong() % (maximum - minimum)) + minimum;

By using the '%' operator, we take the remainder when divided by your maximum value. This leaves us with only numbers from 0 (inclusive) to the divisor (exclusive).

For example:

public long randLong(long min, long max) {
    return (new java.util.Random().nextLong() % (max - min)) + min;
}

Answer 6

The standard method to generate a number (without a utility method) in a range is to just use the double with the range:

long range = 1234567L;
Random r = new Random()
long number = (long)(r.nextDouble()*range);

will give you a long between 0 (inclusive) and range (exclusive). Similarly if you want a number between x and y:

long x = 1234567L;
long y = 23456789L;
Random r = new Random()
long number = x+((long)(r.nextDouble()*(y-x)));

will give you a long from 1234567 (inclusive) through 123456789 (exclusive)

Note: check parentheses, because casting to long has higher priority than multiplication.