Re: Scalars & atomic values & variables

"Lauri Pietarinen" <lauri.pietarinen_at_atbusiness.com> wrote in message news:btck3p$7bt$1_at_nyytiset.pp.htv.fi...
>
> Bob Badour wrote:
>
> >"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 ?!?
> >
> What I meant was that if we decide that the only relational operators
> are the ones that can
> be derived from the primitive ones (or a set of primitives, such as
> UNION, PROJECTION,RESTRICTION)
> then a function that returns a relation from a string is not a
> relational operator. Or it is some kind of hybrid,
> which of course is perfectly OK, but not the kind I had in mind. Would
> that be arbitrary?

Thus, an array is a scalar type because some arbitrary list of operations apply to it but a tuple is non-scalar, but why? Received on Mon Jan 05 2004 - 23:24:30 CET