Oracle FAQ | Your Portal to the Oracle Knowledge Grid |
Home -> Community -> Usenet -> comp.databases.theory -> Re: The wisdom of the object mentors (Was: Searching OO Associations with RDBMS Persistence Models)
Cimode wrote:
[...]
> If a
> fonction is defined as
>
> y(x) = f(x) + p(x)
>
> ...are '=' and '+' functions? No they certainly are not. They are
> simply operators period that help make an equality assumption (which
> sets the equation) based on on an addition operation. Only y(x), f(x)
> and p(x) are functions.
You are terribly confused, amigo. As usually understood in math, in the given context:
'=' stands for 'defined as', it has got nothing to do with the equality relation or some other mysteripous operator. In order to avoid ambiguity and confusion, standard math texts use '=Def' or just 'Def' notatio instead.
'+' is an addition operation which is a synonym for a function with two arguments. It's time to re-read that old secondary school algebra textbook.
>
> <<Operators of type T -> T -> boolean, right?>> boolean can not be
> applied in definition of an operator (for communication purposes, I
> would say "valid and applyable" would be closer to help define what an
> operator does)
This passage does not make any obvious sense. You may want to sort out your terminology before engaging in the discussion.
>
> << So you're talking about functions over a single domain, ones of type
> T
> > -> T, or T -> T -> T, etc. How are these not functions?>> Respectfully *you* are talking about functions not me. I clearly stated that relational operators are not functions. Your symbology is not familiar to me.
That's probably because your comp. science prof has confused you with the pretty formulas like T->T-> that do not appear to have any meaning. The old school textbook migh (or might not) re-acquaint you with words like 'operation', 'function', 'relation', etc.
>
> << So you're talking about the algebraic definition of a type, like
> pop(push(S, x)) = x for a stack?>>No. See above
>
> erk wrote:
> > Cimode wrote:
> > > I am sorry but operators are *not* functions (relation model speaking).
> >
> > What do you mean, "relation model speaking"? Relational operators, or
> > operators on various types (the values of which can be stored in the
> > attributes of tuples of relations)? Or are you using the term
> > "relational" more loosely in the sense that =, <, >=, etc. are
> > relational operators? Operators of type T -> T -> boolean, right?
> >
> > > They are a set of symbols (generally mathematical such as =) that can
> > > be applied to permissible values included in a specific domain.
> >
> > So you're talking about functions over a single domain, ones of type T
> > -> T, or T -> T -> T, etc. How are these not functions?
> >
> > > Each specific combination of operators applied help define a data type.
> > > That's all there's to it.
> >
> > So you're talking about the algebraic definition of a type, like
> > pop(push(S, x)) = x for a stack?
I am afraid the poster has a very vague idea of what he is trying to talk about.
> >
> > - Eric
Received on Fri Jun 02 2006 - 18:29:00 CDT