[wplug] umask

[Sal Mangiapane]

> -----Original Message-----
> From: wplug-admin at wplug.org [mailto:wplug-admin at wplug.org]On Behalf Of
> Mark Dalrymple
> Sent: Tuesday, April 29, 2003 2:11 PM
> To: wplug at wplug.org
> Subject: Re: [wplug] umask
> 
> 
> > I don't remember what operation is used between the 
> > permissions and the umask to get the actually permissions used. umask is 
> > part of bash, so you could look at man bash for more.
> 
> The umask is used when creating files.  The system takes the 
> permissions the application asks for, and then removes the bits
> specified in the umask.
> 
> So, if the app asks for
> 
>  u   g  o     (user group other)
> rwx rwx rwx   (read write execute)
> 111 111 111   777 (octal)
> 
> and the umask is
> 000 010 010   022 (octal)
> 
> The resulting result is
> 111 101 101
> or,
> rwx r-x r-x
> 
> (for the geeks, the umask is complemented, the bitwise ANDed with the
> permissions the program asks for)

If this is bitwise ANDed it would be:
000 010 010
or,
--- -w- -w-
And when you do chmod that is exactly what happens.  So, chmod must be bitwise ANDed.


For me, I like to think of it as the binary value:
124 124 124 this seems easier for me.  a 7 is 1+2+4 (111).  a 5 is 1+0+4 (101). a 644 is 011 001 001 or -wx --x --x

I'm not sure, but UMASK must be bitwise ORed which gives the 111 101 101 result that Mark described.

> 
> Put another way, 022 umask will strip off the write permissions for
> group and for others for newly created files.  umask has not affect on
> existing files.  (if aprogram saves a file by creating a new one and
> writing to it, then the umask will apply to that)
> 
> ++Mark Dalrymple, markd at badgertronics.com.  http://badgertronics.com
>    "apathy! apathy is our cry A-P-aaa forget it."
>     -- jo2y
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> 

Thanks and God bless,

sal