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 [email protected] with a subject of "unsubscribe". Trouble? Contact [email protected] Archive: https://lists.debian.org/[email protected]
