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Date:      Sun, 08 Jun 1997 05:29:29 +1000
From:      David Nugent <davidn@unique.usn.blaze.net.au>
To:        Steve Howe <un_x@anchorage.net>, freebsd-hackers@freebsd.org
Subject:   Re: signed/unsigned cpp 
Message-ID:  <199706071929.FAA28336@unique.usn.blaze.net.au>
In-Reply-To: Your message of "Mon, 02 Jun 1997 22:21:16 PST." <Pine.BSF.3.95q.970602221732.5317A-100000@aak.anchorage.net> 

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> From owner-freebsd-hackers@FreeBSD.ORG  Tue Jun  3 17:16:57 1997
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> Date: Mon, 2 Jun 1997 22:21:16 -0800 (AKDT)
> From: Steve Howe <un_x@anchorage.net>
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> To: freebsd-hackers <hackers@FreeBSD.ORG>
> Subject: Re: signed/unsigned cpp
> In-Reply-To: <199706030619.PAA01871@genesis.atrad.adelaide.edu.au>
> Message-ID: <Pine.BSF.3.95q.970602221732.5317A-100000@aak.anchorage.net>
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> 
> On Tue, 3 Jun 1997, Michael Smith wrote:
> 
> > No, there you are quite right.  But if I use a "char *" type, I know
> > not to explicitly expect it to be either "signed" or "unsigned".
> 
> i'm in the middle of porting a bunch of code, and would like to do
> it as "properly" as possible.  can anyone tell me an instance
> where declaring "char *" is of any benefit, as opposed to
> explicitly defining "unsinged char *" or "signed char *" ?

Backwards compatibility.

Some code doesn't care about the signedness of whatever it is
pointing at, and it simply won't have any impact. Only the fact
that it is pointing at an object who's sizeof() is guaranteed to
be 1 may matter, for the sake of doing pointer arithmetic, for
example.

In c++, the concept of "default signedness" is still there simply
because it makes it possible to more easily compile existing C
code in C++ mode. The addition of "signed" as an explicit type
assists in tighter type checking.

Regards,
David

David Nugent - Unique Computing Pty Ltd - Melbourne, Australia
Voice +61-3-9791-9547  Data/BBS +61-3-9792-3507  3:632/348@fidonet
davidn@freebsd.org davidn@blaze.net.au http://www.blaze.net.au/~davidn/



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