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?

score 27 · Accepted Answer · answered Dec 05 '12 at 14:33

27

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));

answered Dec 05 '12 at 14:33

Yogendra Singh

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+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. – Peter Lawrey Dec 05 '12 at 14:38
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Why didn't you use nextLong()? – St.Antario Nov 21 '15 at 07:36

COME FROM · Answer 2 · 2016-06-15T08:05:52.800

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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.

edited Jun 15 '16 at 08:05

answered Dec 05 '12 at 17:27

COME FROM

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This is not correct; it will return values larger than n on occasion. – Stefan Reich Jun 14 '16 at 22:27
@StefanReich True. Low bits of n will be ignored. I'll edit the answer. – COME FROM Jun 15 '16 at 07:26
@StefanReich I removed the bad example. – COME FROM Jun 15 '16 at 08:06

Java: Random long value in an interval

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