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  1. #1
    Just Joined!
    Join Date
    Jan 2005
    Bangalore India

    A query of umask

    Hi Forum,
    Well I have a doubt on "Umask". Umask is supposed to set the file permissions in Linux when a file automatically created. When I type umask I get a octal number 0022. I have read that umask is suppose to be a 3 digit octal number. But when I type umask at CLI, I get 0022 which is a 4 digit octal number. Could u plz give me a detail description about to how umask is dervied. When the book "Running Linux" has specified the umask to be 3 digit octal number where did four digits come here. Please explain.?????

    Thanks in Advance

  2. #2
    Linux Engineer
    Join Date
    Nov 2004
    Montreal, Canada
    0022 right meens this.

    022 is the "normal" 3 digit right and the first digit(0) in the occurence, is the group ownership right.

    Hope you get the point, dont hesitate to ask any further question
    \"Meditative mind\'s is like a vast ocean... whatever strikes the surface, the bottom stays calm\" - Dalai Lama
    \"Competition ultimatly comes down to one thing... a loser and a winner.\" - Ugo Deschamps

  3. #3
    Linux Engineer
    Join Date
    Nov 2002
    Queens, NY
    Well, it's really 0022.

    The first octet controlls things like SUID, GUID, and sticky bit. Do a search and see what you can find. If you still have questions, feel free to post them here.
    The best things in life are free.

  4. $spacer_open

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