Re: Inconsistent arithmetic evaluation of parameters

[Dennis Williamson]

On Thu, Sep 3, 2015 at 8:45 AM, Stephane Chazelas <stephane.chazelas@gmail.com> wrote:

2015-09-01 16:23:08 -0400, Greg Wooledge:

> On Tue, Sep 01, 2015 at 03:13:57PM -0500, Dennis Williamson wrote:
> > The version of dash I have handy (0.5.7) has math support which IMHO is
> > broken:
> >
> > $ foo=bar
> > $ bar=5
> > $ echo $foo
> > bar
> > $ echo $((foo))
> > dash: 4: Illegal number: bar
> > $ echo $(($foo))
> > 5
> > $ echo $((bar))
> > 5
> > $ echo $(($bar))
> > 5
> >
> > Note the inconsistency in support of omitting the inner dollar sign.
>
> $foo is expanded to bar, so the following two lines are always going to
> be equivalent:
>
> echo $(($foo))
> echo $((bar))
>
> POSIX also specifies (vaguely!!) that $((x)) and $(($x)) are equivalent.

Note that while POSIX may (vaguely indeed) say $((x)) and
$(($x)) are equivalent at least when x contains a litteral
number, $((-x)) and $((-$x)) are not equivalent in all shells if
$x contains a negative number.

$ a=0 x=-1 bash -c 'echo $((a-$x))'
bash: a--1: syntax error in _expression_ (error token is "1")
$ a=0 x=-1 bash -c 'echo $((a-x))'
1

(they're OK in shells that don't implement the (optional in
posix) -- and ++ operators like dash.

Or you can add spaces to make it more reliable: $((a - $x))

In ksh/zsh/bash, see also the difference between:

$ a=1+1; echo $((a*2))
4
$ a=1+1; echo $(($a*2))
3

--
Stephane
>
>

Also, for completeness:

$ a=0 x=-1 bash -c 'echo $(($a-$x))'

1

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