Re: arithmetic operator precedence

[Vincent Lefevre <vincent@vinc17.org>]

On 2008-06-17 14:20:39 +0100, Stephane Chazelas wrote:
> 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))

Yes, but not more, and only if x is a valid integer constant.
Note that recognizing negative constants is not required by POSIX;
so, they are extensions, just like "1+1".

BTW, I've just sent a mail to the Austin group about that and other
things related to arithmetic expansion.

> 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).

No, POSIX doesn't say that. This sentence is a mean to define the
value of x from the contents of $x. But note that parsing has already
been done and you have something like a C expression ("The arithmetic
expression shall be processed according to the rules given in
Arithmetic Precision and Operations"); variables are just replaced
by their values, like in C. Without any extension, both interpretations
are equivalent anyway.

> 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)).

But if you want to use extensions and/or more complex expressions,
you should add parentheses, e.g. x=$((($x) + 1)). This will also
work.

> > 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.

Not completely. The shell may recognize "1 + 1" as an integer constant
as an extension, in which case 6 would be the only valid result. Note
that negative numbers are in the same case.

Also ** is out of the scope of POSIX too.

> No, POSIX does say that $((a ** 2)) is the same as $(($a ** 2))
> because $a contains an integer constant, and that's $((-3**2)).

No (see above). You are 3 times wrong. POSIX doesn't say anything
about **, POSIX doesn't say that non-trivial expressions $((a ...))
and $(($a ...)) are equivalent (and in pratice, they aren't), POSIX
doesn't say anything about negative constants (probably a bug, but
the whole section needs to be improved anyway). Perhaps instead of
saying falsehood, you should read the standards.

-- 
Vincent Lefèvre <vincent@vinc17.org> - Web: <http://www.vinc17.org/>
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Work: CR INRIA - computer arithmetic / Arenaire project (LIP, ENS-Lyon)