Re: Inconsistent arithmetic evaluation of parameters

[Greg Wooledge]

On Wed, Sep 02, 2015 at 11:24:42AM -0400, Chet Ramey wrote:
> On 9/2/15 11:19 AM, Greg Wooledge wrote:
> > On Wed, Sep 02, 2015 at 10:16:14AM -0500, Dennis Williamson wrote:
> >> The $ is implied.
> > 
> > That is completely absurd.  (And wrong.)
> 
> Not exactly.  When the arithmetic evaluator encounters a token that is of
> the form of a shell identifier (`bar'), in a context where an operand is
> needed, it treats it as a shell variable and looks up the variable's
> value.  In that sense, it's an expansion.
> 
> The difference between bash and dash is what each shell does with that
> value.

$foo and foo are not equivalent in dash, as we've already discussed:

$ dash
$ foo=bar bar=5
$ echo $((foo))
dash: 4: Illegal number: bar
$ echo $(($foo))
5

They are also not equivalent in bash, if I may be permitted to change
the context slightly:

$ bash
$ foo='$(id >&2)'
$ echo $((x[foo]))
bash: $(id >&2): syntax error: operand expected (error token is "$(id >&2)")
$ echo $((x[$foo]))
uid=563(wooledg) gid=22(pgmr) groups=1002(webauth),208(opgmr)
0

So "The $ is implied" is false, in both shells, although there are some
contexts where they happen to produce the same result.