I didn't misunderstand. I just find it confusing that $foo.pick
returns $foo when $foo isn't a collection. It's logical but
surprising, orthogonal instead of diagonal. I'd rather see .pick fail
for non-collections.
I get that Bool.pick is supposed to work either way.
On Tuesday, June 29, 2010, Carl Mäsak via RT
<perl6-bugs-follo...@perl.org> wrote:
> On Mon Jun 28 11:38:07 2010, markjreed wrote:
>> That is terribly confusing. I'm tempted to argue for .pick not being
>> defined on scalar values.
>>
>> Do any of the other non-Enumerated types besides Bool have a finite
>> range? I'd say .pick should return an appropriate random value when
>> applied to any such.
>
> I think you misunderstand.
>
> The confusion doesn't originate in Perl 6, but in Rakudo. It wouldn't
> manifest if Bool were
> implemented as an enumeration type.
>
> Bool is an enumerated type in the spec. By the spec, .pick on an enumeration
> type *should*
> already return an appropriate random value. So far in Rakudo, Bool hasn't
> been implemented
> as an enumeration type. Therefore (and only therefore) are you seeing this
> result. Arguing for
> .pick not being defined on scalar values doesn't make any sense.
>
> If it's really really important that Bool.pick return a random Bool, you can
> always inject your
> own .pick method in Bool to do this:
>
> use MONKEY_TYPING;
> augment class Bool {
> method pick() {
> defined self ?? self !! (True, False).pick
> }
> }
>
> In fact, I see no reason not to patch Rakudo itself with this method for the
> time being.
>
--
Mark J. Reed <markjr...@gmail.com>
