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

Question

Starting from Java 7 (or Android API Level 21 = 5.0+) you could directly use ThreadLocalRandom.current().nextLong(n) (for 0 ≤ x < n) and ThreadLocalRandom.current().nextLong(m, n) (for m ≤ x < n). See @Alex's answer for detail.

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If you are stuck with Java 6 (or Android 4.x) you need to use an external library (e.g. org.apache.commons.math3.random.RandomDataGenerator.getRandomGenerator().nextLong(0, n-1), see @mawaldne's answer), or implement your own nextLong(n).

According to https://docs.oracle.com/javase/1.5.0/docs/api/java/util/Random.html nextInt is implemented as

 public int nextInt(int n) {
     if (n<=0)
                throw new IllegalArgumentException("n must be positive");

     if ((n & -n) == n)  // i.e., n is a power of 2
         return (int)((n * (long)next(31)) >> 31);

     int bits, val;
     do {
         bits = next(31);
         val = bits % n;
     } while(bits - val + (n-1) < 0);
     return val;
 }

So we may modify this to perform nextLong:

long nextLong(Random rng, long n) {
   // error checking and 2^x checking removed for simplicity.
   long bits, val;
   do {
      bits = (rng.nextLong() << 1) >>> 1;
      val = bits % n;
   } while (bits-val+(n-1) < 0L);
   return val;
}

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.

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.

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)

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;
}

Further improving kennytm's answer: A subclass implementation taking the actual implementation in Java 8 into account would be:

public class MyRandom extends Random {
  public long nextLong(long bound) {
    if (bound <= 0) {
      throw new IllegalArgumentException("bound must be positive");
    }

    long r = nextLong() & Long.MAX_VALUE;
    long m = bound - 1L;
    if ((bound & m) == 0) { // i.e., bound is a power of 2
      r = (bound * r) >> (Long.SIZE - 1);
    } else {
      for (long u = r; u - (r = u % bound) + m < 0L; u = nextLong() & Long.MAX_VALUE);
    }
    return r;
  }
}