Explaining local -a 'xkeys=("${!'"$1"'[@]}")'

Question

This code is given as a workaround for array indirection at bash-hackers.org wiki page as an example to get array length and indices indirectly.

It's also mentioned in this Stackoverflow question.

I would like to understand specifically why local -a 'xkeys=("${!'"$1"'[@]}")' works. What is happening here. I understand that there are three different strings to the right of the equal sign:

'xkeys=("${!'
"$1"
'[@]}")'

Why does that work?

This is the code:

isSubset() {
local -a 'xkeys=("${!'"$1"'[@]}")' 'ykeys=("${!'"$2"'[@]}")'
set -- "${@/%/[key]}"

(( ${#xkeys[@]} <= ${#ykeys[@]} )) || return 1

local key
for key in "${xkeys[@]}"; do
    [[ ${!2+_} && ${!1} == ${!2} ]] || return 1
done
}

Thanks

I just explained a bunch of the operation of `isSubset` [here](http://stackoverflow.com/a/26499268/258523) though not specifically this bit. — Etan Reisner, Oct 22 '14 at 02:09

score 1 · Answer 1 · answered Jun 04 '14 at 04:15

1

${!"$1"} fetches the variable named in $1. ${variable[@]} expands to all the elements in the array variable. So this basically copies the array named by $1 into xkeys.

answered Jun 04 '14 at 04:15

tripleee

175,061
34
275
318

I understand that that is what it's supposed to be. I guess the question is why does it work this specific way but not when I put it as just one string (i.e. remove the single quotes). For more context you can see some of my explorations [here](http://ahmed.amayem.com/bash-indirect-expansion-exploration/#accessing_array_length_indices) – amayem Jun 20 '14 at 15:45
The `!` isn't swallowed as part of the expansion of `$1` it is array index expansion (as opposed to array value expansion). The overloaded nature of `!` in this position is complicated and confusing. – Etan Reisner Oct 22 '14 at 02:10

score 1 · Accepted Answer · edited May 23 '17 at 12:34

@tripleee is essentially right, though one detail was missed (and I made the same mistake initially a few minutes ago when writing my explanation of the rest of isSubset in my answer here).

The crucial detail here is that local (and declare and typeset) do their own parsing of their arguments.

Let's walk through the various expansions of this line

local -a 'xkeys=("${!'"$1"'[@]}")' 'ykeys=("${!'"$2"'[@]}")'

Splitting the quoted pieces for now will aid in understanding. That gets us

local -a 'xkeys=("${!' "$1" '[@]}")' 'ykeys=("${!' "$2" '[@]}")'

Which expands to

local -a 'xkeys=("${!' "a" '[@]}")' 'ykeys=("${!' "b" '[@]}")'

Which reassembles into

local -a 'xkeys=("${!'"a"'[@]}")' 'ykeys=("${!'"a"'[@]}")'

Combining adjacent quoted words gets us

local -a 'xkeys=("${!a[@]}")' 'ykeys=("${!a[@]}")'

If this was anything but local we would be done at this point (ok eval also).

For example echo: e() { echo 'xkeys=("${!'"$1"'[@]}")' 'ykeys=("${!'"$2"'[@]}")'; }; e a b outputs xkeys=("${!a[@]}") ykeys=("${!b[@]}")

But since this is local which, as I said before, does its own parsing (and in fact you know it does already because otherwise the {0..5} expansions in the assignment to a wouldn't have worked. Try echo {0..5} vs. "{0..5}" to see what I mean) the single-quoted strings get re-evaluated by local and the array index expansions occur. So ${!a[@]} expands to the indices of a and ${!b[@]} expands to the indices of b.

Explaining local -a 'xkeys=("${!'"$1"'[@]}")'

2 Answers2