Hoping is clear.

EXPLANATION

As pointed out in the answers: umask's value is not mathematically subtracted from default directory and file's permissions.

The operation effectively involved is a combination of AND (&) and NOT (!) boolean operators. Given:

R = resulting permissions
D = default permissions
U = current umask

R = D & !U

For example:

 666& !0053 = 110 110 110 & !000 101 011 110 110 110 & 111 010 100 = 110 010 100 = 624 = rw--w-r-- 

 777& !0022 = 111 111 111 & !000 010 010 111 111 111 & 111 101 101 = 111 101 101 = 755 = rwxr--xr-x 

TIP

An easy way to quickly know the resulting permissions (at least it helped me) is to think that we can use just 3 decimal values:

r = 100 = 4 w = 010 = 2 x = 001 = 1 

Permissions will be a combination of these 3 values.
" " is used to indicate that the relative permission is not given.

666 = 4+2+" " 4+2+" " 4+2+" " = rw rw rw 

So if my current umask is 0053 I know I'm removing read and execution (4+1) permission from group and write and execution (2+1) from other resulting in

 4+2 " "+2+" " 4+" "+" " = 624 = rw--w-r-- 

(group and other already hadn't execution permission)

Hoping is clear.

I'm trying to understand better permissions, so I'm doing some "exercises". Here a sequence of commands that I'm using with their respective output:

$ umask 0022 $ touch file1 $ ls -l file1 -rw-r--r-- 1 user group 0 Mar 16 12:55 file1 $ mkdir dir1 $ ls -ld dir1 drwxr-xr-x 2 user group 4096 Mar 16 12:55 dir1 

That makes sense because we know that default file permission are 666 (rw-rw-rw-) and directories default permissions are 777 (rwxrwxrwx). If I subtract the umask value from these default permissions I have 666-022=644, rw-r--r--, for the file1, so it's coherent with the previous output; 777-022=755, rwx-r-x-r-x, for the dir1, also coherent.

But if I change the umask from 022 to 021 it isn't anymore.

Here is the example for the file:

$ umask 0021 $ touch file2 $ ls -l file2 -rw-r--rw- user group 0 Mar 16 13:33 file2 

-rw-r--rw- is 646 but it should be 666-021=645. So it doesn't work accordingly to the previous computation.

Here is the example for the directory:

$ touch dir2 $ ls -ld dir2 drwxr-xrw- 2 user group 4096 Mar 16 13:35 dir2 

drwxr-xrw- is 756, 777-021=756. So in this case the result is coherent with the previous computation.

I've read the man but I haven't found nothing about this behaviour.

Can somebody explain me why?

I'm trying to understand permissions better, so I'm doing some "exercises". Here's a sequence of commands that I'm using with their respective output:

$ umask 0022 $ touch file1 $ ls -l file1 -rw-r--r-- 1 user group 0 Mar 16 12:55 file1 $ mkdir dir1 $ ls -ld dir1 drwxr-xr-x 2 user group 4096 Mar 16 12:55 dir1 

That makes sense because we know that the default file permissions are 666 (rw-rw-rw-) and directories default permissions are 777 (rwxrwxrwx). If I subtract the umask value from these default permissions I have 666-022=644, rw-r--r--, for the file1, so it's coherent with the previous output; 777-022=755, rwx-r-x-r-x, for the dir1, also coherent.

But if I change the umask from 022 to 021 it isn't any more.

Here is the example for the file:

$ umask 0021 $ touch file2 $ ls -l file2 -rw-r--rw- user group 0 Mar 16 13:33 file2 

-rw-r--rw- is 646 but it should be 666-021=645. So it doesn't work according to the previous computation.

Here is the example for the directory:

$ touch dir2 $ ls -ld dir2 drwxr-xrw- 2 user group 4096 Mar 16 13:35 dir2 

drwxr-xrw- is 756, 777-021=756. So in this case the result is coherent with the previous computation.

I've read the man but I haven't found anything about this behaviour.

Can somebody explain why?