Hi kre, > case "$#" in > 0) set -- cur;; > esac
If this is bash(1) then
(($#)) || set cur
also works. ((...)) is arithmetic evaluation.
--
Cheers, Ralph.
Hi kre, > case "$#" in > 0) set -- cur;; > esac
If this is bash(1) then
(($#)) || set cur
also works. ((...)) is arithmetic evaluation.
--
Cheers, Ralph.