Sender:	rich AT phekda DOT freeserve DOT co DOT uk
Message-ID:	<3E2DBF23.2EF2702@phekda.freeserve.co.uk>
Date:	Tue, 21 Jan 2003 21:44:03 +0000
From:	Richard Dawe <rich AT phekda DOT freeserve DOT co DOT uk>
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To:	djgpp-workers AT delorie DOT com
Subject:	Re: size_t and ssize_t
References:	<200106192104 DOT XAA18859 AT father DOT ludd DOT luth DOT se> <3E2D52C7 DOT 70E1DDEB AT phekda DOT freeserve DOT co DOT uk> <2110-Tue21Jan2003210627+0200-eliz AT is DOT elta DOT co DOT il>
Reply-To:	djgpp-workers AT delorie DOT com

Hello.

Eli Zaretskii wrote:
[snip]
> Sorry, I don't get it: %z is defined to print size_t, not ssize_t,
> right?  If so, why did you expect this snippet to compile without a
> warning?  Does some standard say that ssize_t is interchangeable with
> size_t in this context?

In order: no, yes, yes. Here's what C99 says about "z" in section 7.19.7.1:

"z Specifies that a following d, i, o, u, x,or X conversion specifier applies
to a size_t or the corresponding signed integer type argument; or that a
following n conversion specifier applies to a pointer to a signed integer type
corresponding to size_t argument."

I think that ssize_t should be a signed integer type corresponding to size_t.
So I think %zd would display a ssize_t (or equivalent), %zu would display a
size_t (or equivalent), etc.

With our current definition of ssize_t, gcc generates a warning, because
size_t is a long int, but ssize_t is an int.

Bye, Rich =]

-- 
Richard Dawe [ http://www.phekda.freeserve.co.uk/richdawe/ ]

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