Around line 477, it explains that
our $count;
method ^count { return $count }
Such a I<metaclass method> is always delegated to the C<HOW>
object just as methods like C<.does> are, so it's possible to call
this as C<Dog.count> or C<$dog.count>.
However, around line 1983 it illustrates
But C<Any> gives you shortcuts to those:
$obj.can("bark")
$obj.does(Dog)
$obj.isa(Mammal)
...In general, C<Any> will delegate only those metamethods
that read well when reasoning about an individual object.
Infrastructural methods like C<.^methods> and C<.^attributes>
are not delegated, so C<$obj.methods> fails.
That is, metamethods are not automatically delegated by the dispatcher, but
rather ordinary methods forward to the metamethod to enable that form of calling.
---
Over and over the document warns that you really shouldn't use the shortcut
form. So I'd suggest that the second way is right. If you want a shortcut to
be available (if the method reads well when reasoning about an individual
object) then write one, too.
Or, why define the ^foo form in the first place? As it explains earlier, any
method can be called on the protoobject as long as it doesn't try and access the
undefined parts of the instance. So if you want to call $obj.count, then write
method count, not method ^count.
You can also have a trait that generates the shortcut method as well, as
syntactic sugar. That does not require any new language features.
--John
