Date: Sat, 14 Dec 2002 15:11:41 +0200 From: Ruslan Ermilov <ru@FreeBSD.ORG> To: Bruce Evans <bde@zeta.org.au> Cc: "Andrey A. Chernov" <ache@nagual.pp.ru>, bwk@bell-labs.com, obrien@FreeBSD.ORG, current@FreeBSD.ORG Subject: Re: New AWK bug with collating Message-ID: <20021214131141.GD6946@sunbay.com> In-Reply-To: <20021214202522.L5768-100000@gamplex.bde.org> References: <20021213150942.GE86638@sunbay.com> <20021214202522.L5768-100000@gamplex.bde.org>
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On Sat, Dec 14, 2002 at 09:02:40PM +1100, Bruce Evans wrote:
> On Fri, 13 Dec 2002, Ruslan Ermilov wrote:
>
> > On Fri, Dec 13, 2002 at 04:41:06PM +0300, Andrey A. Chernov wrote:
> > > On Fri, Dec 13, 2002 at 14:32:40 +0200, Ruslan Ermilov wrote:
> > > > Pardon my ignorance here, but the following fragment
> > > > returns -1, doesn't it?
> > > >
> > > > #include <stdio.h>
> > > > void
> > > > main(void)
> > > > {
> > > > int i;
> > > >
> > > > i = (unsigned char)1 - (unsigned char)2;
> > > > printf("%d\n", i);
> > > > }
> > >
> > > It very depends on compiler, i.e. does it implements "value preseving" or
> > > "unsigned preserving" for 'char' type conversions. Or ANSI C vs. common C
> > > mode. Better be safe for both.
> > >
> > > Read 6.10.1.1 section here:
> > > http://wwwrsphysse.anu.edu.au/doc/DUhelp/AQTLTBTE/DOCU_067.HTM
>
> For ANSI C, the result of the subtraction only depends on the width
> of unsigned char. If unsigned char has the same width as int, then
> the result is UINT_MAX; otherwise the result is -1. This is an example
> of the brokenness of "value preserving" conversions -- the value is
> as far as possible from being preserved.
>
> Then assignment to "int i" may cause overflow. There is no overflow if
> the RHS is -1. If the RHS is UINT_MAX, then the result of the assignment
> is implementation-defined. The value is is preserved even less than before.
> I think it is usually -0 on 1's complement machines.
>
> So ache's changes is basically a fix for 1's complement machines. I don't
> see much point in it, sincw we assume 2's complement in most places in
> libc/string (except strcoll() :-). E.g., memcmp() just subtracts the
> unsigned char's and assume that all the conversions turn out like they
> do on 2's complement machines. We actually use an assembler version of
> memcmp on most arches but...
>
Hmm, then how you could explain the difference between -traditional
and -ansi outputs for the following fragment on i386:
int printf(char *, ...);
int
main(void)
{
long long l;
unsigned char c1 = 1;
unsigned char c2 = 2;
l = c1 - c2;
printf("%lld\n", l);
l = -1;
printf("%lld\n", l);
}
Or the same code but with `long' on sparc64.
> > This is handled by the -traditional flag of gcc(1):
> >
> > : `-traditional'
> > :
> > : Attempt to support some aspects of traditional C compilers.
> > : Specifically:
> > :
> > [...]
> > :
> > : * Integer types `unsigned short' and `unsigned char' promote to
> > : `unsigned int'.
> >
> > With -traditional, the code I quoted still produces -1.
>
> It produces overflow which normally gives -1 on 2's complement machines.
>
> > In any case, this section doesn't apply to this case because
> > no conversion described in section 6.10 is ever done here,
> > since both operands are of the same type, "unsigned char".
>
> Yes it does. The common type (for arithmetic operators like subtraction)
> is never smaller than int. Both of the unsigned char operands get
> converted to int in the simplest case where unsigned char is smaller
> than int. See 6.10.1 (5) and 6.10.1.1 about "integral promotions".
>
I stand corrected, thanks for explanations, now I see they do.
Cheers,
--
Ruslan Ermilov Sysadmin and DBA,
ru@sunbay.com Sunbay Software AG,
ru@FreeBSD.org FreeBSD committer,
+380.652.512.251 Simferopol, Ukraine
http://www.FreeBSD.org The Power To Serve
http://www.oracle.com Enabling The Information Age
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