On Fri Nov 27 09:22:20 2015, elizabeth wrote: > So implemented with 61ea661d8dfc04acabf1eb46c > > > Liz > > > On 27 Nov 2015, at 17:17, Patrick R. Michaud <pmich...@pobox.com> > > wrote: > > > > The standard meaning of ".roll" is to randomly select elements > > from a list. So I'd expect .roll on a Range to first convert the > > Range to a list of values and then select from those. > > > > If the intent is to select from the values 0.1, 0.2, 0.3, I'd expect > > the programmer to write: > > > > (0.1, 0.2, 0.3).roll(6) > > > > If the intent is to select from a range of values incrementing by > > 0.1, then: > > > > (0.1, 0.2 ... 10.1).roll(6) > > > > If the intent is to generate six random Num values from a Range, > > then it should probably be (I suspect this isn't implemented yet): > > > > (0.1 .. 0.3).rand xx 6 > > > > Pm > > > > > > On Thu, Nov 26, 2015 at 03:36:49PM +0100, Wenzel P. P. Peppmeyer > > wrote: > >> > >> On Thu, 26 Nov 2015, Elizabeth Mattijsen via RT wrote: > >> > >>>> (0.1 .. 0.3).roll(10).say; > >> > >>> What did you expect? a selection of 0.1, 0.2, 0.3 ?? or 10 random > >>> values between 0.1 and 0.3 inclusive? > >> > >> I would (naive) expect 10x a value between 0.1 and 0.3 . Analog to: > >> > >> (0.1, 0.2, 0.3).roll(10).say; > >> # OUTPUT«(0.3 0.2 0.1 0.2 0.1 0.2 0.3 0.2 0.2 0.1)» > >> > >> However, S03 is quite clear how Range is iterating. > >> > >> 0.1.succ == 1.1; > >> > >> So incrementing by 0.1 can't work. It may be reasonable to fail as > >> early as > >> possible for Range.roll on any Range that is neither Int nor Str on > >> both end > >> points. > >> > >> mfgwp > >
17:37 [Coke] lizmat: you last touched https://rt.perl.org/Ticket/Display.html?id=126664 - is that closable now? 18:31 < lizmat> [Coke]: probably needs tests still marking testneeded. -- Will "Coke" Coleda
