Re: umask?

[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

 



Hi,

> The simple explanation is that the value of umask will be subtracted from 
> the protection mask of whatever you create. If you are creating an 
> executable file and your umask is 022 then the resulting protection on the 
> file will be 777-022==755
> 

Subtracting umask from default permissions will work in many cases, but
not always. If you have a bit set in umask which is reset in
corresponding position in default permissions, you'll get wrong results
subtracting umask value from default permission value. (for example, if
you have 6 in default permission and 3 in umask, the resulting
permission will be 4, not 3).
To get the right results in all cases, you must calculate default
permission value AND (NOT umask value).

[]'s
Marcelo


[Index of Archives]     [Current Fedora Users]     [Fedora Desktop]     [Fedora SELinux]     [Yosemite News]     [Yosemite Photos]     [KDE Users]     [Fedora Tools]     [Fedora Docs]

  Powered by Linux