RE: ada vs. cpp

["Steven Deller" <deller@smsail.com>]

> -----Original Message-----
> From: comp.lang.ada-admin@ada.eu.org
> [mailto:comp.lang.ada-admin@ada.eu.org]On Behalf Of Lutz Donnerhacke
> Sent: Wednesday, October 10, 2001 7:21 AM
> To: comp.lang.ada@ada.eu.org
> Subject: Re: ada vs. cpp
>
> * Dmitry Kazakov wrote:
> >1. If you mean that Ada has no type to represent the whole set of all
> >integers, yes it is true. [as well as it is for all other languages
> >designed for finite discrete machines]
>
> man Haskell. (or any other lazy evaluation language with
> unlimited arithmetics)

Of course that is limited by the machine memory, so it is unlimited, but not
infinite and thus doesn't cover all integers either.

Numerous unlimited arithmetic packages have been defined in Ada for
integers, for rational's and for floating point.  Any type derived from
those packages have the unlimited aspect you are looking for.  And they are
used "naturally", e.g.

  with unlimited_integers;
  use unlimited_integers ; -- I personally think this is justified for
arithmetic packages
  procedure test is
     a,b,c : unlimited_integer ;
  begin
     a := unlimited_integer(100);  -- a := 100; -- if mixed arith defined in
the package
     b := unlimited_integer(100);
     c := a**b ;  -- not your ordinary integer :-)
     a := c**c ;  -- likely to overflow machine memory :-(
  exception
     when storage_error =>
         -- code to handle machine limits
  end test ;

I also believe one could create a symbolic processing type, which just
collected operators without evaluation, and doing symbolic simplification
before assignments.  True lazy evaluation.  Hmmm -- might be interesting to
define, since there would have to be symbolic literals.

Regards,
Steve
"Then, after a second or so, nothing continued to happen."
Steven Deller        Smooth Sailing LLC
410 757 6924         deller@smsail.com