Re: suggestion: shell option for echo to not interpret any argument as an option

[Kerin Millar]

On Wed, 26 Jul 2023, at 1:42 PM, Zachary Santer wrote:
> bash's echo command is broken - YouTube
> <https://www.youtube.com/watch?v=lq98MM2ogBk>
>
> To restate what's in the video, you can't safely use echo to print the
> contents of a variable that could be arbitrary, because the variable could
> consist entirely of '-n', '-e', or '-E', and '--' is not interpreted as the
> end of options, but rather, something to print.
>
> I recognized this and replaced all of my calls to echo with printf some
> time ago.
>
> If POSIX mandates that '--' not be taken as the end of options, then the
> safe thing would be to simply not have echo take any options. Obviously,
> that would break backwards compatibility, so you'd want this to be optional
> behavior that the shell programmer can enable if desired.

echo() { local IFS=' '; printf '%s\n' "$*"; }

-- 
Kerin Millar