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Re: abstract class

From: Brian McBride <bwm@hplb.hpl.hp.com>
Date: Fri, 24 Jan 2003 10:29:59 +0000
Message-Id: <>
To: marc@jfcarrion.com, marc@jfcarrion.com, www-rdf-comments@w3.org

At 01:27 24/01/2003 -0800, Marc Carrion wrote:

> >
>    Yes and No. Let me explain, and I beg your pardon
>again for my english. I understand that any instance
>of a class that is subclass of c is instance of c (The
>same in UML, Java, C++, ...) but I cannot say in a RDF
>instance that (x -rdf:type-> c) (the same in UML,
>Java, C++, ...) I'm going to infer it from the
>instance and the schema. Does it make sense?

Now you want to constrain the abstract syntax so that graphs of the 
following form

   x rdf:type c .
   c rdf:type rdfs:Abstract .

Are illegal.  If you do that, then you can't make the inference you say you 
want to make.

Maybe if you were to explain what, in practical terms in the system you are 
building, you need to accomplish, then we could figure something out.

For example, lets say you were building a tool that accepted data from a 
user and its got an adaptive user interface.  So it reads a schema file and 
then uses that to offer the user choices.
Now lets say you want to ensure that the user interface does not offer the 
user the chance to say that something is an instance of one of these union 
classes you call abstract.  You don't need anything in the standard for 
that.  You just define your own notion of an abstract class and go ahead 
and use it.

>  I'm not
>trying to express a NOT I trying to make a diference
>in the range  property of the Property rdf:type. Is it
>more clear in this way?


> >
> > >   If anything identified with a RDF URI it's a
> > >Resource, why all the classes I define should
> > extend
> > >from resource?
> >
> > I don't know what you mean by "extend from".
>   Excuse me for my english again (Now that I read that
>I don't know why I wrote it) I meant The classes you
>define in the schema should be subclass of
>rdf:Resource, at least all them are in the schema of
>the RDFS.

Because everything that RDF can talk about is a resource.  Therefore all 
members of a class must be a resource.  Therefore that class must be a 
subclass of rdfs:Resource.

Received on Friday, 24 January 2003 05:28:49 UTC

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