Random.nextLong(long bound)?
Vitaly Davidovich
vitalyd at gmail.com
Wed Oct 18 22:11:03 UTC 2017
On Wed, Oct 18, 2017 at 5:54 PM Steven Schlansker <
stevenschlansker at gmail.com> wrote:
> My coworker is implementing an algorithm for which he needs a bounded
> random long.
> He needs to be in full control of the seed and not share the instance
> (i.e. not use ThreadLocalRandom).
>
> Random helpfully provides a nextInt(int bound) but nextLong has no such
> overload.
>
> The code to do so is not super involved, but you only have to briefly
> peruse
> the wasteland of StackOverflow answers to see what happens when you get
> developers
> to solve this problem themselves:
>
>
> https://stackoverflow.com/questions/2546078/java-random-long-number-in-0-x-n-range
>
> Seems like a nextLong(bound) would be a trivial addition with outsize
> benefits.
> Any reason not to add it?
>
> PS --
> nextLong says
> "Returns the next pseudorandom, uniformly distributed {@code long} value
> from this random number generator's sequence."
> and
> "Because class {@code Random} uses a seed with only 48 bits, this
> algorithm will not return all possible {@code long} values."
>
> How can this be true? Given a long L1 which can be generated and L2 which
> cannot, it seems trivial that P(L1) != P(L2)
> and therefore it is not a uniformly distributed long value. Am I
> misunderstanding this concept?
It’s uniformly distributed over the *sequence* that the PRNG produces -
it’s just that the sequence doesn’t produce all long values.
>
>
> Thanks!
> Steven
>
> --
Sent from my phone
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