Java: Random long value in an interval

Question

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

I want to generate a random long value in an interval but it seems that the Random class nextLong() doesn't accept arguments like nextInt(). What can I do here?

Solution 1

If you want range based long values then do the below:

 long LOWER_RANGE = 0; //assign lower range value
 long UPPER_RANGE = 1000000; //assign upper range value
 Random random = new Random();


 long randomValue = LOWER_RANGE + 
                           (long)(random.nextDouble()*(UPPER_RANGE - LOWER_RANGE));

Solution 2

You could use nextInt to generate higher and lower ints of the long. It's even possible to extend the Random-class with your own nextLong-method (even though composition could be a safer choice for more serious programming).

Take a look at the Javadoc of nextInt(int n). A nextLong method could be implemented using the same algorithm. Getting it right could prove a bit tricky. Prepare to do some math with pen and paper. Using a proven library is wise if you're not just coding for fun.

Comments

[Vishy]

+1 Its worth noting that Random uses a 48-bit seed so it won't generate all possible double or long. SecureRandom is slow but will generate all possible values.

[St.Antario]

Why didn't you use nextLong()?

[Stefan Reich]

This is not correct; it will return values larger than n on occasion.

[COME FROM]

@StefanReich True. Low bits of n will be ignored. I'll edit the answer.

[COME FROM]

@StefanReich I removed the bad example.