Date: Sun, 22 Jun 2003 23:40:10 -0700 (PDT) From: des@des.no (Dag-Erling =?iso-8859-1?q?Sm=F8rgrav?=) To: freebsd-standards@FreeBSD.org Subject: Re: standards/52972: /bin/sh arithmetic not POSIX compliant Message-ID: <200306230640.h5N6eAt4035319@freefall.freebsd.org>
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The following reply was made to PR standards/52972; it has been noted by GNATS.
From: des@des.no (Dag-Erling =?iso-8859-1?q?Sm=F8rgrav?=)
To: Wartan Hachaturow <wart@tepkom.ru>
Cc: freebsd-standards@FreeBSD.org, freebsd-gnats-submit@freebsd.org
Subject: Re: standards/52972: /bin/sh arithmetic not POSIX compliant
Date: Mon, 23 Jun 2003 08:32:55 +0200
Wartan Hachaturow <wart@tepkom.ru> writes:
> The following reply was made to PR standards/52972; it has been noted by =
GNATS.
> On Sun, Jun 22, 2003 at 10:43:27PM +0200, Jens Schweikhardt wrote:
> > The recursive processing requires that $(($a+1)) needs to undergo
> > parameter expansion within $(()).
>=20=20
> Right, but this construction works in /bin/sh:
>=20=20
> wart@mojo:~$ /bin/sh
> $ a=3D1
> $ echo $(($a+1))
> 2
Yes. It expands to $((1+1)) which evaluates to 2.
> You've said the problem was with the variable without leading $, like
> this:
> $ a=3D1
> $ echo $((a+1))=20
> arith: syntax error: "a+1"
This *should* work, but doesn't.
> And, as far as my English allows me to judge, the quoted part of SUS
> says that "a" in this construct should be left in output as is (since it
> doesn't have leading $, ${, $(, etc.), shouldn't it?
Yes, it should be left as-is so the part of the code that evaluates
arithmetic expressions knows what variable is involved. For instance,
"$(($a+=3D1)) would expand to "$((1+=3D1))" before evaluation, which makes
no sense, while "$((a+=3D1))" clearly says to increase a with 1.
DES
--=20
Dag-Erling Sm=F8rgrav - des@des.no
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