From: Stephane Chazelas <Stephane_Chazelas@yahoo.fr>
To: Zsh hackers list <zsh-workers@sunsite.dk>,
Richard Hartmann <richih.mailinglist@gmail.com>,
Peter Stephenson <p.w.stephenson@ntlworld.com>
Subject: Re: arithmetic operator precedence
Date: Tue, 17 Jun 2008 14:20:39 +0100 [thread overview]
Message-ID: <20080617132039.GK5016@sc.homeunix.net> (raw)
In-Reply-To: <20080617130246.GL10734@prunille.vinc17.org>
On Tue, Jun 17, 2008 at 03:02:46PM +0200, Vincent Lefevre wrote:
[...]
> It is unspecified. So, the shell is right to choose how it sees it.
>
> > If $a contains an integer constant such as -3, then as per
> > POSIX, $((a * 3)) should be the same as $(($a * 3)), that is
> > $((-3 * 3)).
>
> No, POSIX does not say that. It happens to be the same thing here
> just because of the properties of *, but you can't deduce anything
> for extensions.
It says $((x)) is meant to be the same as $(($x)) which I
understand as any occurrance of a variable name (other than $-,
$?, $0... obviously) in $((...)) should be the same as if the $
was not ommited (when $x contains an integer constant).
That's a recent addition to the text. Only recent versions of
ash (BSD shs) support that for instance, which is why it has
been recommanded for a while to write it $(($x + 1)) instead of
$((x + 1)) in POSIX scripts. And I still use x=$(($x + 1)).
For newer ashes, $((x + 1)) is now accepted and note that you do
get an error in ash if $x is not an integer constant.
> And in practice, shells don't treat $((a * 3)) and
> $(($a * 3)) in the same way:
>
> vin:~> a="1 + 1"
> vin:~> echo $((a * 3))
> 6
> vin:~> echo $(($a * 3))
> 4
But here, $a doesn't contain an integer constant, that's out of
the scope of POSIX.
>
> > If we extend that to the non-POSIX **, that would be:
> >
> > a=-3; $((a ** 2)) should be the same as $((-3 ** 2)).
>
> No, if we extend POSIX in an intuitive way (see above), $((a ** 2))
> should be the same as $(((-3) ** 2)), hence 9. So, no problem with
> $((-3**2)) being -9.
[...]
No, POSIX does say that $((a ** 2)) is the same as $(($a ** 2))
because $a contains an integer constant, and that's $((-3**2)).
--
Stéphane
next prev parent reply other threads:[~2008-06-17 13:20 UTC|newest]
Thread overview: 45+ messages / expand[flat|nested] mbox.gz Atom feed top
2008-06-12 9:57 Stephane Chazelas
2008-06-12 13:12 ` Mikael Magnusson
2008-06-12 13:40 ` Peter Stephenson
2008-06-12 14:47 ` Bart Schaefer
2008-06-12 15:01 ` Stephane Chazelas
2008-06-16 8:17 ` Vincent Lefevre
2008-06-16 8:07 ` Vincent Lefevre
2008-06-16 13:42 ` Peter Stephenson
2008-06-16 13:59 ` Stephane Chazelas
2008-06-16 14:33 ` Vincent Lefevre
2008-06-17 9:19 ` Richard Hartmann
2008-06-17 9:45 ` Stephane Chazelas
2008-06-17 10:24 ` Richard Hartmann
2008-06-17 10:24 ` Richard Hartmann
2008-06-17 10:38 ` Stephane Chazelas
2008-06-17 10:43 ` Peter Stephenson
2008-06-17 11:28 ` Vincent Lefevre
2008-06-17 11:46 ` Peter Stephenson
2008-06-17 12:05 ` Vincent Lefevre
2008-06-19 9:37 ` Jun T.
2008-06-19 9:54 ` Stephane Chazelas
2008-06-19 16:00 ` Vincent Lefevre
2008-06-19 16:20 ` Stephane Chazelas
2008-06-19 17:14 ` Vincent Lefevre
2008-06-19 9:58 ` Peter Stephenson
2008-06-19 12:29 ` Richard Hartmann
2008-06-19 16:04 ` Vincent Lefevre
2008-06-19 16:10 ` Mikael Magnusson
2008-06-19 16:27 ` Stephane Chazelas
2008-06-19 17:25 ` Vincent Lefevre
2008-06-19 17:20 ` Vincent Lefevre
2008-06-17 10:45 ` Richard Hartmann
2008-06-17 11:38 ` Vincent Lefevre
2008-06-17 11:19 ` Vincent Lefevre
2008-06-17 11:57 ` Stephane Chazelas
2008-06-17 12:35 ` Vincent Lefevre
2008-06-17 12:46 ` Stephane Chazelas
2008-06-17 13:02 ` Vincent Lefevre
2008-06-17 13:20 ` Stephane Chazelas [this message]
2008-06-17 14:33 ` Vincent Lefevre
2008-06-17 14:53 ` Stephane Chazelas
2008-06-17 15:49 ` Vincent Lefevre
2008-06-17 14:35 ` Stephane Chazelas
2008-06-17 15:05 ` Vincent Lefevre
2008-06-17 10:54 ` Vincent Lefevre
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