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\         


        

Answer 1

The dollar sign is required in some contexts but not the others:

$ bash --version
GNU bash, version 4.3.42(1)-release (x86_64-redhat-linux-gnu)
Copyright (C) 2013 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

$ echo ((1+2))
bash: syntax error near unexpected token `('

$ echo $((1+2))
3

$ for ((x=0; x<3;++x)); do echo $x; done
0
1
2

$ for $((x=0; x<3;++x)); do echo $x; done
bash: `$((x=0; x<3;++x))': not a valid identifier

---

After reading bash man page, the dollar sign is not required in compound commands:

Compound commands are the shell programming constructs. Each construct begins with a reserved word or control operator and is terminated by a corresponding reserved word or operator. Any redirections (see Redirections) associated with a compound command apply to all commands within that compound command unless explicitly overridden.

In most cases a list of commands in a compound command’s description may be separated from the rest of the command by one or more newlines, and may be followed by a newline in place of a semicolon.

Bash provides looping constructs, conditional commands, and mechanisms to group commands and execute them as a unit.

for (( expr1 ; expr2 ; expr3 )) is a compound command and hence the dollar sign is not required to enable arithmetic evaluation.

Whereas echo $((expr)) is not a compound command because it does not start with a reserved bash keyword, so it requires a dollar sign to enable arithmetic evaluation.

Answer 2

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")'