bash - Arithmetic expansion in array indices - is the dollar sign needed?

Question

When I use arithmetic expansion in an array index in bash, like this:

declare -a FILES
declare -i INDEX=0

for FILE in ./*
do
    FILES[((INDEX++))]="$FILE"
done

Do I need a dollar sign in front of ((...))?
So does it have to be:

FILES[((INDEX++))]="$FILE"

or rather:

FILES[$((INDEX++))]="$FILE"

?

In my local copy of bash both variants seem to work - it's version 4.3.30.

I would expect only the latter to work, because I think only that one returns the result of the arithmetic expression. But there: Bash - arithmetic in array index I've read that only the first might work with older versions of bash (?). So which one is actually correct? And why is the first one working? I haven't found a specific answer to that yet.

Answer 1

In arrays, bash considers expressions between []as arithmetic. Thus

i=2 ; f[i++]=10

is perfect. Writing f[((i++))] is also correct but in this case, (()) is not seen as the arithmetic expansion operator, but as nested parentheses.

Note that ((expr)) evaluates expr, then succeeds if it is true, while$((expr)) is expanded as its value. So f[$((i++))] is also correct.

Finally, f[$i++] is not what you want since $i is expanded first. For instance, i=j ; f[$i++] will be expanded as f[j++].

Remark: a strange feature is that bash expands all it can in arithmetic mode without the $ sign:

$ unset i j k f
$ i=j ; j=k ; k=5 ; f[i++]=10
$ declare -p i j k f
declare -- i="6"
declare -- j="k"
declare -- k="5"
declare -a f='([5]="10")'