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.