Date: Mon, 30 Jun 2025 10:44:24 -0400
From: Greg Wooledge <g...@wooledge.org>
Message-ID: <20250630144424.gj32...@wooledge.org>
| Unfortunately, that's not true. You need $ if you want to use any
| special parameters in an arithmetic context, such as $# or $1.
There are a zillion other reasons why it can be needed as well,
One is to have a variable operator in the expression (say op=+ or op=-
and then use ${op} as a binary operator.)
Further I often embed command substitutions within arithmetic, that's
useful as well (and unless the results would be different using the old
form cmdsub (`...`) which it doesn't seem to be) this really should be
fixed (I personally don't care much about the (()) command, but the
same issue arises in $(( )) -- though that's not surprising, the two
should be the same for everything except the final result use).
And even more, it should be possible to have the entire expression
be the result of a variable expansion
((${exp}))
and (aside from no recursive expansions) that should evaluate exactly
the same as
eval "((${exp}))"
would.
What's more, I don't see what the issue is, inside (( )) or $(( ))
is what should be treated as a double quoted string (arith context
for evaluation, but still a double quoted string - and the first thing
the shell should be doing with any double quoted string, in any context,
is to do any var/cmdsub/arith expansions that occur within it. Then
it gets used for whatever purpose it has.
Do that in the given example, and
((i=$x,a[b[i]]))
turns into
((i=0,a[b[i]]))
which does work. Of course, I know nothing about how any of this
is implemented inside bash, but this just seems simple to me, and
if as:
chet.ra...@case.edu said:
| This was one of the results of this 2021 discussion:
| https://lists.gnu.org/archive/html/bug-bash/2021-03/msg00056.html
| You can get the result you want by not using the $.
then something went very badly wrong with what happened there, and
simply saying "don't use the $" is not nearly good enough.
kre
