printf in Zsh does not shell escape exclamation mark

Question

In Zsh 5.9, we can use printf to shell escape a string:

$ printf '%q' 'One! Two'

One\!\ Two

This produces the correct output of escaping the ! and the space. Now let’s make it as a script:

#!/bin/zsh

printf '%q' "${1}"

Now if we run it, the ! is not escaped but the space is:

$ ./my-script 'One! Two'

One!\ Two

If I change the script to /bin/bash (version 3.2), it correctly escapes the ! and space.

Is this a Zsh bug, or is there some subtle detail I’m missing?

Answer 1

It looks like it checks the current shell settings to determine if the bang is special. History expansion isn't enabled by default in noninteractive shells, so it's not special in a script.

% (setopt   banghist; printf '%q\n' 'One! Two')
One\!\ Two
% (setopt nobanghist; printf '%q\n' 'One! Two')
One!\ Two

The same appears to apply to the ${(q)var} parameter expansion.

Honestly, I find that a somewhat odd choice, as while it means the output is suitable as input to the same shell session, it's not generally suitable as input to every zsh session. Always escaping the bang shouldn't cause any issues as \! gives a ! regardless of if banghist is set.

You could use ${(qq)var} instead to single-quote the string (or ${(qqqq)var} to use the $'...' quotes). The bang would then be quoted along with everything else. Or make it into a function:

% printq() { printf '%s\n' "${(qq)1}" }
% printq one\!two
'one!two'