Iain M Conochie writes: > However: > > $: which umask > $: > > So umask is _not_ a program (in the sense that there is no binary > called umask on the system)
zsh, however, is more helpful: $ which umask umask: shell built-in command Alexis. -- To UNSUBSCRIBE, email to debian-user-requ...@lists.debian.org with a subject of "unsubscribe". Trouble? Contact listmas...@lists.debian.org Archive: https://lists.debian.org/877fzduh3v....@gmail.com
