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Related to file and directory permissions and umask: Why the command umask (shows the octal form) and umask -S (shows the symbolic form) show different (actually the option -S shows the bits negations of the octal form)? I was expecting the umask in symbolic form for umask 024 to be u=---,g=w,o=r but it shows like this:

[user1@server-base ~]$ umask
0024
[user1@server-base ~]$ umask -S
u=rwx,g=rx,o=wx

Terminal window

YAZ84
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  • Partially. I mean it explains the use of umask and how the default permissions are obtained from base/initial permissions and umask. But why umask -S output shows the negation of umask output. – YAZ84 Dec 20 '20 at 09:50
  • This [answer](https://askubuntu.com/a/783144/986805) explains it best. –  Dec 20 '20 at 10:16
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    @YAZ84 It's not the negation but the resulting permissions. Go with your example: `umask 024` `777 - 024` results in `753` which is `u=rwx,g=rx,o=wx`. – mook765 Dec 20 '20 at 10:24
  • [source](https://github.com/bminor/bash/blob/8868edaf2250e09c4e9a1c75ffe3274f28f38581/builtins/umask.def#L146) says: *Print the umask in a symbolic form. In the output, a letter is printed if the corresponding bit is clear in the umask.* so it's seems to be by design. –  Dec 20 '20 at 10:31

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