How would I create this function to choose a random variable from an array?

Question

Here is my attempt:

rand_var() {
printf "%s\n" "${${!1}[RANDOM % ${#${!1}[@]}]}"
}

And I run it like this

array=("something" "somethingelse" "test")
rand_var array

However, it tells me bad substitution. I think it may have something to do with the variables and quoting but I can't figure it out, I use ${!1} so that it acctually uses the contents of the variable and not just array. This line has worked before when I specify the variable name instead of ${!1}.

Related, maybe dupe: [Indirect return of all elements in an array](http://unix.stackexchange.com/q/20171). The short version is that you can't use indirection that way and probably need `eval`. — Michael Homer, Jan 08 '16 at 01:40
Why not just `printf '%s\n' "${array[RANDOM%${#array[@]}]}"` — cuonglm, Jan 08 '16 at 01:49
Actually `nm="$1[@]";tmp=("${!nm}");printf "%s\n" "${tmp[RANDOM % ${#tmp[@]}}"` works without eval, but only by copying the array first. — Michael Homer, Jan 08 '16 at 01:50

score 1 · Answer 1 · answered Jan 08 '16 at 02:16

You can do what you want by copying the array into a local variable, and then selecting from that array:

rand_var() {
    local nm="$1[@]"
    local tmp=("${!nm}")
    printf "%s\n" "${tmp[RANDOM % ${#tmp[@]}]}"
}

I don't think you can do it otherwise, because you can't indirect to the length of the array. If you passed in the length to the function as well:

rand_var_len() {
    local nm="$1[$((RANDOM % $2))]"
    printf "%s\n" "${!nm}"
}
rand_var_len array ${#array[@]}

then you could do it without any copying.

To avoid both, you'll need to use one of the eval-alikes.

How would I create this function to choose a random variable from an array?

1 Answers1