${parameter:?word} expansion for array range

Question

${parameter:?word} prints the given error message if a variable is unset:

$ echo ${unset_var?badness}
bash: unset_var: badness

How to use this on an array?

$ array=(a b c)
$ echo "${array[@]:1:2}"
b c

With:

$ echo "${array[@]:1:2?Need at least two arguments}"

bash complains:

bash: array[@]: a: expression recursion level exceeded (error token is "a")

zsh complains:

zsh: bad math expression: operator expected at `a least tw...'

---

This also "works" differently with positional parameters:

$ echo ${1?Argument not given}
bash: 1: Argument not given

We can print a range in an array, eg:

$ set a b c
$ echo ${@:2}
b c

Or, alternatively:

$ echo ${@:2:$#-1}
b c

bash:

$ set a b c
$ echo ${@:2?badness}
bash: @: 2?badness: `:' expected for conditional expression (error token is "badness")
$ echo ${@:2:$#-1?badness}
bash: @: 3-1?badness: `:' expected for conditional expression (error token is "badness")

zsh:

% set a b c
% echo ${@:2?badness}
zsh: bad math expression: ':' expected
% echo ${@:2:$#-1?badness}
zsh: bad math expression: ':' expected

Is there a way that works in both {z,ba}sh?

Answer 1

In zsh, you could do:

printf ' - "%s"\n' "${(@)a[1,4]}" "${a[5]?Need at least 5 elements}"

In bash as well, but only for non-sparse arrays as ${a[@]:0:5} there like in ksh whose array design bash has mostly copied, that gives you the first 5 elements whose index is >= 0, so you may have more than 5 elements in there with the one with index 4 (bearing in mind that array indexes start at 0 instead of 1 in bash) unset, or the one of index 4 set but be the only one set.

a=( "${a[@]}" ) # unsparsify the array
printf ' - "%s"\n' "${a[@]:0:4}" "${a[4]?Need at least 5 elements}"

Or you could do the sensible thing and write it:

(( $#a >= 5 )) || die 'Need at least 5 elements.' # zsh
(( ${#a[@]} >= 5 )) || die 'Need at least 5 elements.' # ksh/bash (or zsh)

Instead of trying to be too fancy.