Re: Scalars & atomic values & variables

"Lauri Pietarinen" <lauri.pietarinen_at_atbusiness.com> wrote in message news:3FF96A65.9010007_at_atbusiness.com...
> Bob Badour wrote:
>
> >"Lauri Pietarinen" <lauri.pietarinen_at_atbusiness.com> wrote in message
> >news:3FF904F7.3060207_at_atbusiness.com...
> >
> >
> >>Dawn M. Wolthuis wrote:
> >>
> >>
> >>
> >>>Therefore, unless everyone is discussing the exact same set of objects
> >>>(which could be defined by a set of types) and operators, the use of
the
> >>>term scalar is inexact and apt to be confusing as different values will
> >>>
> >>>
> >be
> >
> >
> >>>scalar in different models. When the whole idea of the model is to be
> >>>extensible, then the term scalar is also not very useful as adding in
> >>>another function/operator into the space and/or adding in new objects
can
> >>>change what is and is not a scalar.
> >>>
> >>>
> >>>
> >>If we think of scalars in terms of what can be seen by relational
> >>
> >>
> >operators (union, projection, cartesian product, etc...) it would seem
> >
> >
> >>to me that defining scalars as values that cannot be "cracked open" with
> >>
> >>
> >these operators would pretty well do it for me.
> >
> >That strikes me as entirely arbitrary. One can crack open a string using
an
> >operation that returns a relation of position/character tuples. I guess
that
> >makes a string non-scalar.
> >
> Well, I don't know what the definition of a relational operation is, but
> if we decide that such a function
> is not a relational operator then it is not arbitrary.

Um, am I hearing you correctly? If we make an arbitrary decision, then it is not arbitrary ?!? Received on Mon Jan 05 2004 - 21:59:46 CET