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
> > wrote:
> >
> > The standard meaning of ".roll" is to randomly select elements
> > from a list. So I'd expect .roll on a Range to first
So implemented with 61ea661d8dfc04acabf1eb46c
Liz
> On 27 Nov 2015, at 17:17, Patrick R. Michaud 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 thos
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).
On 11/26/2015 03:36 PM, 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
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;
#
> On 17 Nov 2015, at 19:16, Wenzel Peppmeyer (via RT)
> wrote:
>
> # New Ticket Created by Wenzel Peppmeyer
> # Please include the string: [perl #126664]
> # in the subject line of all future correspondence about this issue.
> # https://rt.perl.org/Ticket/Display.html?id=126664 >
>
>
> (0.
# New Ticket Created by Wenzel Peppmeyer
# Please include the string: [perl #126664]
# in the subject line of all future correspondence about this issue.
# https://rt.perl.org/Ticket/Display.html?id=126664 >
(0.1 .. 0.3).roll(10).say;
# OUTPUT«(0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1)»
# I
