Re: detecting no arguments in $@

[Alex fxmbsw7 Ratchev]

i=0
while (( ++i <= $# )) ; do
 arg=${!i}
done

or

declare -n arg=args[i]
args=( '' "$@" ) i=0
while [[ -v args[++i] ]] ; do
 printf 'arg: %s\n' "$arg"
done

On Sun, Jul 25, 2021 at 1:16 PM <dora-solomon@brusseler.com> wrote:
>
> From: Andreas Kusalananda Kähäri <andreas.kahari@abc.se>
> To: dora-solomon@brusseler.com
> Subject: Re: detecting no arguments in $@
> Date: 25/07/2021 13:11:14 Europe/Paris
> Cc: help-bash@gnu.org
>
> On Sun, Jul 25, 2021 at 01:07:53PM +0200, dora-solomon@brusseler.com wrote:
> >
> > Is it possible to figure out whether there are no arguments in $@ ?
>
>
> If there are no positional parameters, i.e. if "$@" is empty, then "$#"
> would be zero:
>
> if [ "$#" -eq 0 ]; then
> echo 'there were no arguments given'
> fi
>
>
>
> I was using shift  to get passed some options and if the last arguments did 
> not
>
> contain anymore information such as directory names, I would set defaults.
>
>
>
>
>
>