expansions upon arithmetic evaluation in array subscripts

[Stephane Chazelas]

2014-10-13 09:02:38 -0400, Chet Ramey:
[...]
> You have to put it together.  A variable need not be expanded before
> arithmetic evaluation, so the evaluator expands, for instance, a bare
> `a' to `b[$(echo 1+1)]'.  When a variable is expanded, its value is
> treated as an expression to be evaluated.  An array subscript -- for an
> indexed array -- is an arithmetic expression.  All tokens in an arithmetic
> expression, including an array subscript, undergo variable expansion, word
> splitting, and quote removal.
[...]

Thanks. I'm still confused though.

In

b=2
a='1 + $b'
echo $((a))

a is expanded and the expansion is also evaluated as an
arithmetic expression, so why isn't $b expanded there?

Note that it's not only variable expansion, it's also tilde
(even though ~ is also an arithmetic operator) expansion.

$ HOME=1 a='b[~]'  bash -c 'b=(1 2 3); echo $((a))'
2

That means for instance that 

foo=-1
echo $((a[~foo]))

won't work on systems where there's a "foo" user.

It's also command substitution obviously.

Not process substitution though contrary to what I previously
said (which would have been a bigger problem for things like:

arr[x<(a+b) ? x : y]
)

I wasn't aware of word splitting. I didn't think that could
apply given that we're not in list context (splitting into
several words doesn't make sense where). That makes it a
difference with other shells.

$ a='1-+1' c='b[$a]' bash -c 'b=(1 2 3); IFS=-; echo $((c))'
3
$ a='1-+1' c='b[$a]' ksh -c 'b=(1 2 3); IFS=-; echo $((c))'
1

Again, I'd say it's not very consistent and doesn't make much sense:

$ a='1-+1' c='b[$a]' bash -c 'b=(1 2 3); IFS=-; echo $((c))'
3

$ a='1-+1' bash -c 'b=(1 2 3); IFS=-; echo $((b[$a]))'
1

Why splitting (and joinging with space?) in the first case and
not the second?

-- 
Stephane