Date:	Tue, 21 Jan 2003 21:06:27 +0300
From:	"Eli Zaretskii" <eliz AT is DOT elta DOT co DOT il>
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To:	djgpp-workers AT delorie DOT com
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In-reply-to:	<3E2D52C7.70E1DDEB@phekda.freeserve.co.uk> (message from Richard
	Dawe on Tue, 21 Jan 2003 14:01:43 +0000)
Subject:	Re: size_t and ssize_t
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> Date: Tue, 21 Jan 2003 14:01:43 +0000
> From: Richard Dawe <rich AT phekda DOT freeserve DOT co DOT uk>
> 
> Well, we still seem to have ssize_t defined as an int and size_t defined as a
> long. I'm currently coding up support for the 'z' conversion qualifier for
> printf. This specifies that the following integer is of size_t or some
> equivalent sized thing. This is the kind of code that should work IMHO:
> 
>     #include <stdio.h>
>     #include <stdlib.h>
>     #include <sys/types.h>
> 
>     int
>     main (void)
>     {
>       ssize_t foo = 5;
> 
>       printf("%zd\n", foo);
>       return(EXIT_SUCCESS);
>     }
> 
> When I compile it, I get this warning:
> 
> gcc @../../../gcc.opt -I. -I- -I../../../../include  -c foo.c
> cc1.exe: warnings being treated as errors
> foo.c: In function `main':
> foo.c:17: warning: signed size_t format, ssize_t arg (arg 2)
> make.exe: *** [foo.o] Error 1

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?

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