Re: Controlling ('rand') random number upper limit

Oooh, I missed that.

Very nice!  I'll have to remember that one!

randomreturn = ((Max_Range - 1) / Random * Range) + 1

Regards,

Jim Langston

Me transmitte sursum, Caledoni!

Chris Rehm wrote:
> 
>      His message did show that he was using 32768 as the divisor. So,
> he'll get a value between 0/32768 and 32767/32768.
>    This gives me a chance to use the only thing I've ever liked about
> Windows, the calculator it comes with. So he'll get a value ranging from
> 0 to .999969482421875 (barring any typos on my part).
>    That should give as flat a distribution across his own range as the
> generator has across its own range, provided he does not wish to
> generate numbers larger than 32767. If he does, he will need a different
> random number generator or to append multiple results from this
> generator bitwise and then divide by the new range.
> 
> Jim Langston wrote:
> 
> > Chris, Paul,
> >
> > That being said, Paul's solution can still be screwed, depending
> > on how the system is adjusting for math.  One out of 32767 times
> > it will return a value 1 greater than his range.
> >
> > For a range 1 to 10 (normal return):
> > 32766 / 32767 = 0.999969481 * 10 = 9.99969481.  Truncate returns
> 
> [snip] s/b 32766/32768
> 
> > Regards,
> >
> > Jim Langston
> >
> --
> 
>    Chris Rehm
> <mailto:javadisciple@earthlink.net>

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