Re: Inconsistent arithmetic evaluation of parameters

[Dennis Williamson]

On Tue, Sep 1, 2015 at 3:23 PM, Greg Wooledge <wooledg@eeg.ccf.org> wrote:

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))

It's the line above those two where I demonstrate the failure in the indirection and equivalency.

$ echo $((foo))  # expansion succeeds, indirection fails

dash: 4: Illegal number: bar

$ echo $(($foo))  # both expansion and indirection succeed

5

$ echo $((bar))  # expansion succeeds

5

$ echo $(($bar))  # expansion succeeds

5

 

Bash doesn't have this inconsistency. (Just don't omit the dollar sign from $RANDOM in most Bourne-derived shells. https://lists.gnu.org/archive/html/bug-bash/2010-01/msg00039.html)

POSIX also specifies (vaguely!!) that $((x)) and $(($x)) are equivalent.

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