Hawson: Rewrite.

2006-01-12T22:37:15Z

Rewrite.

← Older revision		Revision as of 15:37, 12 January 2006
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	umask is a mask to the default permissions ~~that~~ a file is ~~written~~. ~~Consider this if~~ you write a file its default mode is 0666. This means anyone can ~~delete~~ it, ~~write to it and read from it~~. If you had a umask of 077, ~~the bits would~~ be masked when ~~you write the~~ file. Here is the formula for determining the mode after umask is applied: ~~'''~~mode &~~= ~(~~umask~~)'''~~. ~~Let's look at~~ this ~~in detail what does it mean? mode is mode~~ (bitwise) AND ~~unary inverse~~ of umask (~~every bit set to 0 becomes 1 and vice versa~~). ~~So 0666 AND 0700 is 0600~~, ~~and that is what your final~~ mode ~~is.~~ ~~So a~~ umask of ~~077~~ is ~~very safe, noone but~~ the ~~user gets any~~ permissions~~. What if~~ you want ~~to restrict only writing to a file created~~ by ~~group~~ and ~~others? It would~~ be ~~a umask of 022~~, ~~so 0666 AND 0755~~ and ~~you have 0644~~.		umask is a mask agaist the default [[permissions]] used when a file is created.

			If you write a file, its default mode is 0666. This means anyone can read or write it (actually removing a file requires directory write access, which is independent of the permissions on a specific file.

			If you had a umask of 077, ''all'' permissions for "group" and "other" will be masked (e.g. disallowed) when a new file is created.

			Here is the formula for determining the mode after umask is applied: <code>mode & !umask</code>. In "English," this means that the you take the umask, apply a unary inverse (e.g. a logical <code>NOT</code>), then bitwise <code>AND</code> this value against the default mode of the new file.

			So, given a default mode of 666, and a umask of 027, the following math is performed (behind the scenes!):

			666 & !027 = 640

			Which is <code>rw-r----</code>. Working out the binary, we have this:

			110110110 (default mode 666)
			000010111 (umask of octal 027, in binary)

			Negate the umask:
			!000010111 = 111101000

			Bitwise AND the negated umask against the default mode
			110110110
			& 111101000
			-----------
			110100000 = 640

			Simple, right? ;-)

			Just remember that the umask is used to set the permissions that you want ''prevent'' from getting set by default, and you'll be okay.

			Common mask settings:
			{\|border=2 cellpadding="2"
			\|Umask
			\|Default file permissions
			\|Notes
			\|-
			\|077
			\|600 (rw-------)
			\|Very restrictive, good for root, and paranoid users
			\|-
			\|027
			\|640 (rw-r-----)
			\|Common for users who want to let certain people read files
			\|-
			\|002
			\|660 (rw-rw-r--)
			\|Good for collaboration and active sharing of files.
			\|-
			\|}

Pbug: attempt at explanation of umask

2005-12-03T22:09:47Z

attempt at explanation of umask

New page

umask is a mask to the default permissions that a file is written. Consider this if you write a file its default mode is 0666. This means anyone can delete it, write to it and read from it. If you had a umask of 077, the bits would be masked when you write the file. Here is the formula for determining the mode after umask is applied: '''mode &= ~(umask)'''. Let's look at this in detail what does it mean? mode is mode (bitwise) AND unary inverse of umask (every bit set to 0 becomes 1 and vice versa). So 0666 AND 0700 is 0600, and that is what your final mode is. So a umask of 077 is very safe, noone but the user gets any permissions. What if you want to restrict only writing to a file created by group and others? It would be a umask of 022, so 0666 AND 0755 and you have 0644.

Umask - Revision history

Hawson: Rewrite.

Pbug: attempt at explanation of umask