Re: [vox-tech] how does umask work?

[Foo Lim <fMAPSwlim@ucdavis.edu>]

On Tue, 27 Mar 2001, Peter Jay Salzman wrote:

> On Tue 27 Mar 01, 10:31 AM, Henry House said: 
> > On Tue, Mar 27, 2001 at 10:38:59AM -0800, Peter Jay Salzman wrote:
> > > ok, i give up.  kind of looks like the number is subtracted from 666, but
> > > that's not quite correct.
> > > 
> > > how does this work?  how does one set the x bit, for instance (not that
> > > you'd want to, but i'm just curious).
> > > 
> > > umask 000   # rw-rw-rw-
> > > umask 111   # rw-rw-rw-
> > > umask 222   # r--r--r--
> > > umask 333   # r--r--r--
> > > umask 444   # -w--w--w-
> > > umask 555   # -w--w--w-
> > > umask 666   # ---------
> > > umask 777   # ---------
> > 
> > There are four digits; the first is assumed to be octal zero if omitted.
>  
> are you sure?
> 
> % umask 1111
> bash: umask: `1111' is not an octal number from 000 to 777
> 
> i think you're thinking of chmod, not umask.
> 
> ok, so how do i set a umask of --x--x--x ?
> 
> pete

The umask is subtracted from 777, not 666.  If you created an executable,
then a umask of 111 will give you a 666 permission.  You couldn't see a
difference between 000 & 111 above because you weren't creating an
executable.  I hope this helps!  FL