Zsh: How to make `${x:?}` return non-zero in case of empty `$x`?

Question

How to make ${x:?} return non-zero in case of empty $x? By default, it exits the shell (subshell in this example) with 0.

function sth() {
    local url="${1:?}"
...
}
(sth) ; echo $? # shows 0

Update: As Stéphane Chazelas says, this only happens in interactive usage.

Note that it doesn't just exit the function, it exits the (sub)shell. — Stéphane Chazelas
– Stéphane Chazelas, Commented Mar 12, 2021 at 14:30
Correction, I get 1 in scripts (non-interactive shells), but 0 in interactive shells. That sounds related to zsh.org/mla/users/2016/msg00755.html — Stéphane Chazelas
– Stéphane Chazelas, Commented Mar 12, 2021 at 14:35
That syntax is from the Bourne shell (from the late 70s) which didn't have functions initially. It's mostly useful in scripts outside of functions. — Stéphane Chazelas
– Stéphane Chazelas, Commented Mar 12, 2021 at 16:10

Marlon Richert · Accepted Answer · 2021-03-16 12:34:46Z

-1

As explained in the comments, ${x:?} itself cannot do that. However, what you want can be done easily with a conditional statement:

% sth() { [[ -n $1 ]] || return; local url="$1" }
% sth; print $?
1

answered Mar 16, 2021 at 12:34

Marlon Richert

4,29811 silver badges38 bronze badges

1 Answer 1