[racket] Question about round

I was only advocating it in the case of infinities. The R(5,6)RS
documentation for min/max says: "If any argument is inexact, then the result
is also inexact (unless the procedure can prove that the inaccuracy is not
large enough to affect the result, which is possible only in unusual
implementations)." I would claim that the inaccuracy in the min/max in this
case is finite and not large compared to infinity, so the result is not
affected.

I seem to remember that PLT Scheme used to handle infinities as exact
quantities. Can anyone confirm or deny that?

Doug

On Fri, Sep 30, 2011 at 3:42 PM, Jos Koot <jos.koot at telefonica.net> wrote:

> **
> I already wrote:
>
> #lang racket
> (= (inexact->exact (round (/ 1.0 1.1e-200)))
>  (round (/ #e1.0 #e1.1e-200))) ; -> #f
>
> A computation that ends up with an integer may introduce a computational
> error when during the computation inexact numbers have been used.
> Inexactness should remain contaguous.
>
> I do admit that (min 0 +inf.0) can retuirn an exact integer.
>
> Jos
>
>
>  ------------------------------
> *From:* users-bounces at racket-lang.org [mailto:
> users-bounces at racket-lang.org] *On Behalf Of *Mark Engelberg
> *Sent:* viernes, 30 de septiembre de 2011 18:18
> *To:* Stephen Bloch
> *Cc:* users at racket-lang.org
> *Subject:* Re: [racket] Question about round
>
> I'm amazed at how long I've gone with the misconception that:
> Exact Numbers = Integers union Rationals
> Inexact Numbers = Floating Point numbers
> (Frankly, I never really thought much about what infinities were.  I
> assumed they were a separate, distinct type).
>
> When I saw in the docs for round that it returns an integer, and I was
> getting an inexact number, I was sure it was a mistake.  Thanks for helping
> me straighten out my thinking on this.  I can see how you might want to test
> an inexaxct number for whether it is the floating point approximation of an
> "integer", so that makes sense.  Still seems weird and inconvenient for
> round to give you back an inexact integer rather than an exact one, but I
> can see now how this behavior matches the standard.
>
> Thanks,
>
> Mark
>
>
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