On Sat, Nov 30, 2024 at 10:02:49AM +0000, Klemens Nanni wrote:
> > +
> > +Index: ejabberdctl.template
> > +--- ejabberdctl.template.orig
> > ++++ ejabberdctl.template
> > +@@ -126,7 +126,7 @@ set_dist_client()
> > + exec_cmd()
> > + {
> > + case $EXEC_CMD in
> > +- as_install_user) su -s /bin/sh -c 'exec "$0" "$@"' "$INSTALLUSER"
> > -- "$@" ;;
> > ++ as_install_user) su -s /bin/sh "$INSTALLUSER" -c 'exec "$0" "$@"'
> > "$@" ;;
> > + as_current_user) "$@" ;;
>
> I don't understand why argv is passed twice, the one inside the -c string is
> enough, no?
>
> Isn't $@ the whole command incl. argv[0] aka. $0? If so, 'exec_cmd echo foo'
> would print
> "echo foo" and not "foo" due to the 'exec $0'.
The outer $@ and inner $@ are different, and the one inside -c isn't
expanded on the outer script. The first positional parameter after
`sh -c '...'` will be assigned to $0, and the rest to $@. Passing
`echo foo` will run `echo`.
