Re: What databases have taught me
From: Dmitry A. Kazakov <mailbox_at_dmitry-kazakov.de>
Date: Mon, 3 Jul 2006 14:39:17 +0200
Message-ID: <pz6oon49y1h2$.1x8rq4unalorp$.dlg_at_40tude.net>
>
> AFAIK (given my very restricted mathematical knowledge...), this is the
> definition of an "algebraic structure" :
>
> """
> an algebraic structure consists of one or more sets closed under one or
> more operations, satisfying some axioms.
> """
>
> So is there a 1:1 mapping between CS types and algebraic structures ?
Date: Mon, 3 Jul 2006 14:39:17 +0200
Message-ID: <pz6oon49y1h2$.1x8rq4unalorp$.dlg_at_40tude.net>
On Mon, 03 Jul 2006 11:41:53 +0200, Bruno Desthuilliers wrote:
>> On Fri, 30 Jun 2006 17:37:34 GMT, Jay Dee wrote: >> >>>Well, no. A function, let's say, an operation on integers which >>>returns a rational (approximation, of course), like DIVIDE, >>>requires that the types exist before the function -- but the >>>types don't require the function at all. Granted, many OO >>>languages bundle the methods up in the class -- but that's a >>>mistake. >> >> It is a philosophical question. In case you have missed my previous posts, >> I reject data. So my answer is no. Types don't exist without operations and >> values. All operations have to be defined as well as values.
>
> AFAIK (given my very restricted mathematical knowledge...), this is the
> definition of an "algebraic structure" :
>
> """
> an algebraic structure consists of one or more sets closed under one or
> more operations, satisfying some axioms.
> """
>
> So is there a 1:1 mapping between CS types and algebraic structures ?
If you add semantics to the type (as invariants, preconditions of operations etc), then it probably will.
-- Regards, Dmitry A. Kazakov http://www.dmitry-kazakov.deReceived on Mon Jul 03 2006 - 14:39:17 CEST