topic Re: umask in Operating System - HP-UX

umask

Macho_2 — Thu, 11 Jan 2007 01:34:41 GMT

How to read the umask setting like 022,027,0002 and 0113. Please explain the umask setting in the following format or any other easy way.
rwx--x-wx

Re: umask

Ivan Krastev — Thu, 11 Jan 2007 02:07:58 GMT

This may be useful - http://en.wikipedia.org/wiki/Umask

regards,
ivan

Re: umask

rariasn — Thu, 11 Jan 2007 02:35:01 GMT

Hi,

Sample use "umask".

$ umask 077
$ umask --> 077
$ umask -S --> u=rwx,g=,o=

$ man umask

rgs,

Re: umask

D Block 2 — Thu, 11 Jan 2007 22:09:19 GMT

try this one:
cd /tmp
mkdir fun_umask
cd fun_umask
umask 022
touch file_with_022
ls -l file_with_022

grep umask /etc/profile

what is your systems default umask ?

now try this one:

umask 0
touch file_with_000
ls -l file_with_000

why did HP-UX provide the default umask settings ? A well known security hole, that's why..

the "touch" command provides a simple method of calling the system call function: creat().. using the umask(1) setting from the parent shell.

let me know if you need more lessons here..

best of luck and of couse we expect points, right :-)

Re: umask

Patrick Wallek — Thu, 11 Jan 2007 22:42:20 GMT

First, you must remember the default modes used to create files and directories. The default permissions for a file are -rw-rw-rw- (666) and directories drwxrwxrwx (777).

Now the umask "subtracts" (kinda sorta, but not really, but enough for this explanation) the umask bits from the default permissions.

So:

Example 1:

umask = 022

You create a file:

default perms 666 minus umask 022 = permissions on file of 644 (-rw-r--r--).

You create a directory:

default perms 777 minus umask 022 = permissions of 755 (drwxr-xr-x)

Example 2:

umask = 027

You create a file:
666 - 027 = 640 (-rw-r-----)

Create a dir:

777 - 027 = 750 (drwxr-x---)

If the umask has 4 digits, you can drop the leading 0 (as long as its a 0).

Re: umask

Raj D. — Thu, 11 Jan 2007 22:42:54 GMT

Macho,

For example you have umask set to 022, so when a file is created by default it gets (666-022) permission i.e 642 .

if you hav umask set to 000 , so the default file permission will be 666-000 = 666

umask on file and directory differs,

for
file = (666-umask value)
dir = (777-umask value)

Now the number convention to rwx

r w x
Owner: 400 200 100
Group: 40 20 10
Other: 4 2 1
-------------------------

So Umask 022 will provide default file creation permission to ( 666-022) = 644
i.e -rw-rw----

Hope this will help,
Cheers,

Raj.

Re: umask

D Block 2 — Thu, 11 Jan 2007 22:57:20 GMT

Macho,

7 7 7
111 111 111

rwx rwx rwx

chmod 777 myfile

: provides mode 777

this assumes your umask is set to ZERO or 000

umask 000
touch myfile
ls -l myfile

mode of file is 777 or rwx-rwx-rwx

x = 1
w = 2
r = 4
=====
7

Re: umask

Patrick Wallek — Thu, 11 Jan 2007 22:57:50 GMT

Raj,

Sorry, but NO. 666 - 022 = 644 which = -rw-r--r-- (try it and see!)

The numerical notations are:

(r)ead = 4
(w)rite = 2
e(x)ecute = 1

This is the same for user/owner, group and other/world.

So rwxrwxrwx = 421 421 421 = 777 (if you add each triplet together).

With the chmod command you can use 'chmod 755 file' or 'chmod u=rwx,go=rx file' to grant rwxr-xr-x permissions to a file. u=user (or owner), g=group and o=other (or world). The chmod man page explains this. (I know you didn't ask about chmod, but I'm hoping this will help you understand.)

Re: umask

D Block 2 — Thu, 11 Jan 2007 23:08:58 GMT

Macho, back to my math lesson..

if you have mode:

7 7 7

and umask is set (set to > 000), then you subtract.

example:

$ umask 022

math:
7 7 7
-
0 2 2
=====

well, the first 0 value might be interesting here.. what result do you get ?

try this one:

umask 022
touch file_junk
ls -l file_junk

do you get:
rwx-rx-rx

or

766

given: rwx = 7
rx = 5 (r=4 + x=1) = 5

again: rwx is simply the math of:
r = 4
w = 2
x = 1
=====
total 7

Re: umask

D Block 2 — Thu, 11 Jan 2007 23:18:53 GMT

sorry, my math is a little off

umask 0
touch file
mode is 7 7 7
or
rwx rwx rwx

umask 022
touch file2
mode is 755 <--- note 755

or
rwx r x r x
not group write, not others write

best of luck and thanks for being a member here !!

Re: umask

Raj D. — Thu, 11 Jan 2007 23:27:26 GMT

Patrick,
Thanks for correcting , I mean that only
= -rw-r--r-- , but in a hurry i made the mistake.

644 is = -rw-r--r-- i.e [ ( 400+200+0) + (40+0+0) + (4+0+0) ] so it comes 644.

thx ,
Cheers,
Raj.

Re: umask

Raj D. — Fri, 12 Jan 2007 00:08:55 GMT

Macho,

Also dont forget to check this out,

http://www.cactus2000.de/uk/unit/massunx.shtml

Enjoy,
Raj.

Re: umask

Macho_2 — Fri, 12 Jan 2007 01:25:49 GMT

Dear All;

Thanks for all your help.

Re: umask

Arturo Galbiati — Fri, 12 Jan 2007 08:50:07 GMT

Hi,
umask value has been subtracted (octal value for group user, group other) by the 666 for file file/777 for directory value, so:

# Permissions
# umask (file/directory)
------ -------------------
000 rw-rw-rw-/rwxrwxrwx
002 rw-rw-r--/rwxrwxr-x
022 rw-r--r--/rwxr-xr-x
027 rw-r-----/rwxr-x---
077 rw-------/rwx------

HTH,
Art

Re: umask

Raj D. — Fri, 12 Jan 2007 09:40:45 GMT

Macho,

"I have assigned points to 26 of 38 responses to my questions. "

That means you have not assigned points to 12 replies of those who replied to your thread.

I suggest you to make it 38/38 , to stay in touch with this forum with great responses, and giving value of the posts those who are using their time and trying to help.

Also check out the link:
http://66.34.90.71/ITRCForumsEtiquette/

cheers,
Raj.

Re: umask

Dennis Handly — Sat, 13 Jan 2007 00:00:09 GMT

>Patrick: Now the umask "subtracts" (kinda sorta

The correct C expression is:
int umask = 022;
int perm = 0666;
int result_perm = perm & ~umask;

(It is AND Complement.)

Or in a real shell:
$ typeset -i8 um=8#22 perm=8#666; echo $um $perm
8#22 8#666
$ typeset -i8 final_perm
$ (( final_perm = perm & ~ um ))
$ echo $final_perm
8#644