does this all mean that c++ is ACTUALLY D ?
hummmmmmmm..... food for thought.....
Pierre
----- Original Message -----
From: "Kevin Meltzer" <[EMAIL PROTECTED]>
To: "Paul Johnson" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>; "Nick Transier" <[EMAIL PROTECTED]>;
<[EMAIL PROTECTED]>
Sent: Wednesday, June 27, 2001 4:38 PM
Subject: Re: Incrementing Strings
> On Wed, Jun 27, 2001 at 06:21:30PM +0200, Paul Johnson ([EMAIL PROTECTED])
spew-ed forth:
> > On Wed, Jun 27, 2001 at 11:07:12AM -0400, Kevin Meltzer wrote:
> > >
> > > Is that the reason? I would think --'a' would be z, but that is my own
> > > internal logic :)
> >
> > But ++'z' isn't 'a'.
>
> No, it is aa, the next logical thing to come after z. So, --a could be
> aaaaaa..... or, to me, more logically z. But, I don't scream that my
> logic is always logical ;)
>
> > > But, why is --'a' (or any a-zA-Z) -1? Why doesn't, at least, it
evaluate
> > > 'a' to true (1) and --'a' = 0 (or undef, since I don't know why it
> > > should return a true value)? I guess there must be a reason, but it
> > > isn't documented from what I can see.
> >
> > Converting 'a' to a number gives 0.
>
> I'm not convinced on that. Being that magic is built in to make a++ into
> b.. so it isn't being converted to 0 in that case, which would make that
> statement false. Why would it be 0?
> Why not 1? Why is 'b' also converted to 0? Why not use it's ord() value?
> To me (again, internal, warped, logic) --a returning ` makes more sense
> than -1. z++ is aa, so why isn't aa-- reverted back to z? Why must --a
> be seemingly useless?
>
> Cheers,
> Kevin
>
> --
> [Writing CGI Applications with Perl - http://perlcgi-book.com]
> "Families is where out nation finds hope, where wings take dream."
> -- G.W. Bush, LaCrosse, WI 10/18/2000
