# New Ticket Created by Minimiscience
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- Time::gmtime doesn't handle having an optional argument well:
$ perl6
> sa
On Sun, Jul 12, 2009 at 12:07:14AM -0400, Chas. Owens wrote:
> On Sat, Jul 11, 2009 at 22:02, Minimiscience wrote:
> > - How does one declare multiple variables of the same type with a single
> > "my" statement? Is it "my Int ($x, $y);", "my(Int $x, Int $y);", or
> > something else? Are the pare
On Jul 12, 2009, at 12:07 AM, Chas. Owens wrote:
Since grep is defined as returning a list of matching elements and
first is
defined as being the same as grep, I would say that it returns an
empty list
if nothing matches. The empty list is one of the false values.
Does the empty list count
On Jul 12, 2009, at 9:46 AM, Patrick R. Michaud wrote:
".?method" seems to work for me in Rakudo:
$ cat x
my $x = undef;
say ($x.?foo).perl;
$ ./perl6 x
undef
This doesn't work when the variable is assigned to a typed container:
#!/usr/bin/env perl6
use v6;
my Str
On Sun, Jul 12, 2009 at 9:29 AM, Minimiscience wrote:
> On Jul 12, 2009, at 12:07 AM, Chas. Owens wrote:
>>
>> Since grep is defined as returning a list of matching elements and first
>> is
>> defined as being the same as grep, I would say that it returns an empty
>> list
>> if nothing matches. Th
Em Sex, 2009-07-10 às 15:39 -0700, Jon Lang escreveu:
> The key to understanding roles is to note that roles don't implement
> methods; classes implement methods.
Er, while I see your point, Roles are not just interfaces... they are OO
components that can be plugged into other classes. They often
payload++ brought this up on #perl6:
in current Rakudo, $string ~~ /re/ sets $/ in the scope in which the
expression appears, ie
'a' ~~ /./;
say $/; # ouput: a
But $str.match(..) and $str.subst don't. The spec is rather silent, it
says "There are also method forms of m// and s///: [...]
Em Dom, 2009-07-12 às 22:51 +0200, Moritz Lenz escreveu:
> I setting of OUTER::$/ considered syntactic sugar?
> I don't care either way, I'd just like some clarification so that I can
> write tests and submit tickets (if appropriate).
As far as I remember, it's not really OUTER::$/, but each routi
Minimiscience wrote:
> I tried to find the answers to these in the Synopses, but I couldn't.
> Plan B is to ask the mailing list.
>
> - What does the "first" method/subroutine return when no elements of
> the list match? Does it return the empty list? Does the return value
> count as un
# New Ticket Created by Moritz Lenz
# Please include the string: [perl #67480]
# in the subject line of all future correspondence about this issue.
# http://rt.perl.org/rt3/Ticket/Display.html?id=67480 >
Rakudo: 08928df70e4af4013eb0bfde02256262728ecf6e
$ perl6 -e 'my @a = ; say @a...@a()'
Ca
On 2009-Jul-10, at 4:37 pm, Jon Lang wrote:
This is one of the distinctions between role composition and class
inheritance. With class inheritance, the full tree of inherited
classes is publicly accessible; with role composition, the methods
of the combined role are the only ones that are m
On 2009-Jul-12, at 12:43 pm, Daniel Ruoso wrote:
role R1 {
method foo() {...} # degenerates to interface
}
Just wondering: since merely declaring an interface will be common
enough, should we be able to say simply "method foo;" inside a role,
and drop the {...}?
class Bla does R2 {
m
On Jul 12, 2009, at 20:15 , David Green wrote:
sub nighttime (Canine $rover) { $rover.bark if any(burglars()); }
(...)
3) $rover acts like a Canine, but the rest of the original $dogwood
arg (the Tree parts) are still there; they just aren't used unless
somehow explicitly brought out; for
Daniel Ruoso wrote:
> Jon Lang wrote:
>> The key to understanding roles is to note that roles don't implement
>> methods; classes implement methods.
>
> Er, while I see your point, Roles are not just interfaces... they are OO
> components that can be plugged into other classes. They often are used
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