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Re: Perl 6 Meta Object Protocols and $object.meta.isa(?)
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From:
Stevan Little
Date:
August 8, 2005 14:50
Subject:
Re: Perl 6 Meta Object Protocols and $object.meta.isa(?)
Message ID:
6eda6c7e2a3f905740eea9af6d8fe2b5@iinteractive.com
Mark,
On Aug 8, 2005, at 4:26 PM, Mark Reed wrote:
> Coming in late here, but it seems odd to have an actual class called
> "MetaClass". The meta-object protocols with which I am familiar have
> the
> concept of a metaclass (a class whose instances are themselves
> classes), and
> the class Class is such a metaclass, but where does a class named
> MetaClass
> fit in?
I discussed with Larry at the hackathon about the role that Class
played in the metamodel. We decided that is was but a thin veneer
between the meta-land and the user-land. I assume this is still the
case. Here is a 10,000 ft view of the metamodel prototype I sketched
out the other day
(http://svn.openfoundry.org/pugs/perl5/Perl6-MetaModel/docs/
10_000_ft_view.pod). It should shed a little light on this discussion.
As for how this differs from the other MOPs out there. I took the basic
design of MetaClass, Class, Object from Smalltalk -80 actually, but
modified the relationships a little to be more has-a the is-a. Let me
expand/digress on this slightly ..
The basic Smalltalk-80 idea of every user-level Class having an
associated meta-level Class is still retained. However in Smalltalk,
Class is an instance of MetaClass, where in the metamodel Class has-a
instance of MetaClass. And as I said, Class is really nothing special,
it but a level of indirection between the instance and MetaClass
instance.
I also borrowed many ideas from CLOS (in particular from book "The Art
of the MetaObject Protocol"). CLOS is more like what you describe,
where standard-class is the metaobject to define classes. I see this as
mapping to the MetaClass, and our Class as being something akin to the
find-class generic function in CLOS.
> If all metaclasses are instances of MetaClass, then MetaClass must
> be an instance of itself - is this then the only cycle in the graph?
Yes, that is the cycle.
>
>> 1) MetaClass is a subclass of Object
>> 2) MetaClass is an instance of MetaClass
>
> OK.
>
>> So the following code should be true (given a random instance $obj).
>>
>> $obj.meta.isa(MetaClass);
>> $obj.meta.isa(Object);
>
> What does $obj.meta return - is it just a shortcut for
> $obj.class.class, or
> is something else going on here?
Well, I did not see $obj.class speced in A/S12 so I never did anything
with that. However the p6opaque instance structure I use in the
prototype metamodel has a pointer back to the class object (see the
10,000 ft view again). So it is simple to implement it if we want too.
But to answer your question, I was always under the impression that
$obj.meta returned the MetaClass instance associated with the class
that $obj is an instance of.
However, keep in mind, these are somewhat fuzzy areas in Syn/Apoc12,
and all details about $obj.meta only deal with "Introspection".
<best pirate voice>
Arghhh, these be uncharted waters 'mah Boy!
</best private voice>
>
> If the former, then all of these should be true.
>
> $obj.isa(Object)
yup, this will always be true.
> $obj.class.isa(Object)
> $obj.class.isa(Class)
Again, no .class that I know of, however if there is, then these too
should be true.
> $obj.meta.isa(Object)
> $obj.meta.isa(Class)
> $obj.meta.isa(MetaClass)
I will agree with 1 and 3, but not with 2. I see Class and MetaClass
are seperate things, at least how I coded it. However this, should be
true (assuming we introduce a .class method):
$obj.meta.class.isa(Class)
However this is not a closed issue, so we can discuss it if you see a
real need for things to be this way.
>> However, Syn/Apoc 12 shows that the following is true if $foo is an
>> instance of the Foo class.
>>
>> $foo.meta.isa(Foo)
>
> Hm. That doesn't make sense to me at all. Clearly I need to reread
> the
> Syn/Apoc. I'd expect $foo.isa(Foo) and that's it, although if nothing
> fancy
> with composition is going on $foo.class == Foo would also be true.
Yup, doesn't make sense to me either :)
- Stevan
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