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

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)

Answer 2

According to http://java.sun.com/j2se/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;
}

Answer 3

From the page on Random:

The method nextLong is implemented by class Random as if by:

public long nextLong() {
   return ((long)next(32) << 32) + next(32);
}

Because class Random uses a seed with only 48 bits, this algorithm will not return all possible long values.

So if you want to get a Long, you're already not going to get the full 64 bit range.

I would suggest that if you have a range that falls near a power of 2, you build up the Long as in that snippet, like this:

next(32) + ((long)nextInt(8) << 3)

to get a 35 bit range, for example.