Date: Thu, 16 May 2002 12:11:57 -0700 From: "David O'Brien" <obrien@FreeBSD.ORG> To: Bill Fenner <fenner@research.att.com> Cc: audit@FreeBSD.ORG, current@FreeBSD.ORG Subject: Re: moused(8): char signed-ness problem with gcc 3.1 Message-ID: <20020516121157.F67791@dragon.nuxi.com> In-Reply-To: <200205161906.MAA06850@windsor.research.att.com>; from fenner@research.att.com on Thu, May 16, 2002 at 12:06:49PM -0700 References: <200205160346.UAA27116@windsor.research.att.com> <86k7q48h2w.wl@archon.local.idaemons.org> <20020516114247.A67791@dragon.nuxi.com> <200205161906.MAA06850@windsor.research.att.com>
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On Thu, May 16, 2002 at 12:06:49PM -0700, Bill Fenner wrote:
> >Specifically what is the problem? Given the program below, take the
> >ISO-C spec and explain the problem. Or even w/o the spec -- I haven't
> >been reading this thread.
>
> > > int
> > > main()
> > > {
> > > unsigned char i = 127;
> > > char j;
> > >
> > > printf("%d\n", ((char)(i << 1)));
>
> This prints -2, which is correct -- (signed char)254 is -2.
>
> > > j = ((char)(i << 1)) / 2;
> > > printf("%d\n", j);
>
> This prints 127, which is incorrect. -2 / 2 is -1.
>
> > > j = ((char)(i << 1));
> > > printf("%d\n", j / 2);
>
> This breaks down the previous expression into two halves, and
> results in the correct answer of -1. However, there should
> be no difference between
> ((char)(i << 1)) / 2
> and
> char j = ((char)(i << 1); j / 2
Sounds like you should forward this to gcc-bugs@gcc.gnu.org. :-)
CC audit@ (and current@) so we can all see what the GNU people say.
Don't forget to show the output of `cc -v'.
Or run `gccbug' from a recient gcc31 port install. If you do this
instead, please send me the PR #.
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