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 routine
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
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 {
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
for
