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Old 07-13-2003, 09:03 PM   #1
InsaneBob
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chmod and umask


I need to change permissions for some folders and files but I do not get how the octal number representation work. Also how do you use umask with fstab file?
 
Old 07-13-2003, 09:09 PM   #2
MasterC
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Here's the best I've ever seen on an explanation:
http://www.linuxquestions.org/questi...threadid=25234

Cool
 
Old 07-13-2003, 09:39 PM   #3
Mathieu
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A small tutorial on how to Setup Permissions
http://www.realitywebs.com/unix/permissions.shtml
 
Old 07-13-2003, 10:03 PM   #4
itsjustme
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Here, this was fun. See if it helps. Corrections, more explanations welcome:
(i hope this formatting works.)

Code:
8+4+2+1 = 15 = 8+4+2+1
- - - - - -- - - - - -
0 0 0 0 = 00 = 0+0+0+0
0 0 0 1 = 01 = 0+0+0+1
0 0 1 0 = 02 = 0+0+2+0
0 0 1 1 = 03 = 0+0+2+1
0 1 0 0 = 04 = 0+4+0+0
0 1 0 1 = 05 = 0+4+0+1
0 1 1 0 = 06 = 0+4+2+0
0 1 1 1 = 07 = 0+4+2+1
1 0 0 0 = 08 = 8+0+0+0
1 0 0 1 = 09 = 8+0+0+1
1 0 1 0 = 10 = 8+0+2+0
1 0 1 1 = 11 = 8+0+2+1
1 1 0 0 = 12 = 8+4+0+0
1 1 0 1 = 13 = 8+4+0+1
1 1 1 0 = 14 = 8+4+2+0
1 1 1 1 = 15 = 8+4+2+1

Chop off the '8' part and:

owner group others
4-2-1 4-2-1 4-2-1 (value)
r w x r w x r w x (mode)

Examples:

chmod 444
--4-- --4-- --4--
4+0+0 4+0+0 4+0+0 
1 0 0 1 0 0 1 0 0
r - - r - - r - -

chmod 744
--7-- --4-- --4--
4+2+1 4+0+0 4+0+0
1 1 1 1 0 0 1 0 0
r w x r - - r - -

chmod 665
--6-- --6-- --5--
4+2+0 4+2+0 4+0+1
1 1 0 1 1 0 1 0 1
r w - r w - r - x

etc...
.
Code:
[bsl1@localhost bsl1]$ chmod 444 test.txt
[bsl1@localhost bsl1]$ ls -l test.txt
-r--r--r--    1 bsl1     bsl1           15 Jul 13 21:13 test.txt
[bsl1@localhost bsl1]$ chmod 744 test.txt
[bsl1@localhost bsl1]$ ls -l test.txt
-rwxr--r--    1 bsl1     bsl1           15 Jul 13 21:13 test.txt
[bsl1@localhost bsl1]$ chmod 665 test.txt
[bsl1@localhost bsl1]$ ls -l test.txt
-rw-rw-r-x    1 bsl1     bsl1           15 Jul 13 21:13 test.txt
Code:
umask:
subtract the 'values' in the umask columns from 777
 to get the default permission set for directories 
(not sure about the deal with files):

[bsl1@localhost bsl1]$ umask
0002
[bsl1@localhost bsl1]$ mkdir zzz
[bsl1@localhost bsl1]$ ls -ld zzz
drwxrwxr-x    2 bsl1     bsl1         4096 Jul 13 21:13 zzz
(default)

-----rwx-rwx-rwx
0777 (7) (7) (7)
-----111 111 111 (4+2+1)(4+2+1)(4+2+1)
----------------
0002 (0) (0) (2) (umask)(minus)
----------------
0775 (7) (7) (5)
-----111 111 101 (4+2+1)(4+2+1)(4+0+1)
-----rwx rwx r-x
Isn't this fun!

Last edited by itsjustme; 07-13-2003 at 10:38 PM.
 
Old 07-13-2003, 10:59 PM   #5
itsjustme
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Quote:
Originally posted by Mathieu
A small tutorial on how to Setup Permissions
http://www.realitywebs.com/unix/permissions.shtml
If you can get your mind around the relationship of '4 2 1' with the mode bits 'r w x' as I posted above, then you don't need to do all that adding of 400 200 100 40 20 10 4 2 1.

It's the bit relationships, not the actual value in the hundreds.

regards...
 
Old 07-14-2003, 03:54 PM   #6
itsjustme
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More that I just learned about umask from:

Introduction to Linux
A Hands on Guide
Machtelt Garrels
CoreSequence.com
<tille@coresequence.com>
Version 3.0.2 Last updated 20030228 Edition

"...before the mask is applied, a directory has permissions 777 or rwxrwxrwx, a plain file 666 or
rw−rw−rw−.
The umask value is substracted from these default permissions after the function has created the new file or
directory. Thus, a directory will have permissions of 775 by default, a file 664, if the mask value is (0)002."
 
Old 07-16-2003, 08:21 AM   #7
lido
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Smile

Hi,

I'm using linux the first time and trying to understand the permission.

The permission i have for my newly created directory and file is different. My default umask is 0022. I'm wondering why the umask value shown in my pc is having 4 digits instead of the standard 3 digit 022? And using 777 to minus away 022 should be 755. But both my directory and file does not have the permission of 755 as shown below. Appreciate if anyone can kindly explain to me...

[root@cm]# mkdir directory
[root@cm]# ls -ld directory
drwxr-xr-x 2 root root 4096 Jul 16 20:17 directory


[root@cm]# touch file.c
[root@cm]# ls -ld file.c
-rw-r--r-- 1 root root 0 Jul 16 20:18 file.c
[root@cm]# umask
0022
 
Old 07-16-2003, 08:56 AM   #8
Mons
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Hello itsjustme:
Your small piece of tutorial was wonderful and fun.
Thank you for posting it, and do continue with it.
Mons...
 
Old 07-16-2003, 09:52 AM   #9
JZL240I-U
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Quote:
Originally posted by lido
...
The permission i have for my newly created directory and file is different. My default umask is 0022. I'm wondering why the umask value shown in my pc is having 4 digits instead of the standard 3 digit 022? And using 777 to minus away 022 should be 755. But both my directory and file does not have the permission of 755 as shown below. Appreciate if anyone can kindly explain to me...

[root@cm]# mkdir directory
[root@cm]# ls -ld directory
drwxr-xr-x 2 root root 4096 Jul 16 20:17 directory


[root@cm]# touch file.c
[root@cm]# ls -ld file.c
-rw-r--r-- 1 root root 0 Jul 16 20:18 file.c
[root@cm]# umask
0022
Please read post #6 carefully again .

drwxr-xr-x
d4+2+1, 4+1, 4+1 = 7,5,5 = 7,7,7 - 0,2,2

-rw-r--r--
-4+2, 4, 4 = 6,4,4 = 6,6,6 - 0,2,2

Okay?

Last edited by JZL240I-U; 07-16-2003 at 10:09 AM.
 
Old 07-16-2003, 09:58 AM   #10
itsjustme
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Here is a link to that Introduction to Linux

Go down to "3.4.2.2. The file mask" and check that out. You can ignore the first 0 in 0022 for this.

lido, 755 is rwxr-xr-x .

Touch just changes the file time stamp. What do you get when you create a new file? Should be 0666 - 0022 = 0644 = rw-r--r-- (oh, which is what you do have for file.c).

Directory default before umask = 777
File default, before umask = 666
(according to that Introduction to Linux book.)

(rwx)(r-x)(r-x) = (4+2+1)(4+0+1)(4+0+1) = 755 = 777 - 022
(rw-)(r--)(r--) = (4+2+0)(4+0+0)(4+0+0) = 644 = 666 - 022


I think I'm right, therefore I am.

regards...


Edit: Oh, hey, thanks JZL240I-U .

Last edited by itsjustme; 07-16-2003 at 10:09 AM.
 
Old 07-16-2003, 10:13 AM   #11
JZL240I-U
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Quote:
Originally posted by itsjustme
I think I'm right, therefore I am.
Edit: Oh, hey, thanks JZL240I-U
I like that one. And you are wellcome
 
Old 07-16-2003, 10:42 AM   #12
lido
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Thumbs up

I understand now.

Thank you very much for your explanation. You guys are fantastic!!!!
 
Old 07-16-2003, 11:19 AM   #13
Rumblefish
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Wow...

That's a topic that's really difficult to explain properly...and you guys did an excellent job. I'm very, very impressed!
 
Old 07-16-2003, 11:54 AM   #14
itsjustme
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Re: chmod and umask

OOPS....

Quote:
Originally posted by InsaneBob
Also how do you use umask with fstab file?
For me, I'll have to do some more research to find the answer to that...
 
Old 07-16-2003, 12:31 PM   #15
MasterC
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Umask is just chmod reversed. Count backwards from 7, so for a umask of 7 you would use 0. For 6, you would use 1, and so on:
0 : 7
1 : 6
2 : 5
3 : 4
4 : 3
5 : 2
6 : 1
7 : 0



And as a side note:
Happy Birthday ItsJustMe!

Cool
 
  


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