Shell Programming and Scripting

I'm getting to grips with this concept of the umask.

What I thought was, setting umask uga+rwx would result in creating files with all permissions for everyone. Seems not to be the case though. Read and write bits get set, but not the execute bit.

Is there some gap in my understanding, or is some other setting overwriting the usual behavior of umask?

Demonstration follows:

Code:

$umask uga-rwx; rm a-file; touch a-file; ls -l a-file
rm: remove write-protected regular empty file `a-file'? y
----------  1 hartmann users 0 2006-06-07 21:11 a-file
$umask uga+rwx; rm a-file; touch a-file; ls -l a-file
rm: remove write-protected regular empty file `a-file'? y
-rw-rw-rw-  1 hartmann users 0 2006-06-07 21:11 a-file

Let's use numbers, because umask can be confusing. It works sort of backwards.
umask controls the process' default file protection mask, chmod works on single files.
umask affects every file you create, in other words.

umask 777 looks like it would set all the bits on, but umask works on the "complement",
ie, it does the opposite, so this creates a protection mask with all bits off.

chmod 777 filename plays fair - it sets the bits on.

umask 000 would turn on bits that it could turn on. It works backwards, sort of.

also, all umask does is modify the default behavior of permissions
on directory and file creation.

by default, a directory is created with 777 and a file 666.

so, if you set your umask to 0, meaning, allow the permissions
to be at their highest settings possible during file creation
you still don't get files to be created as executables by default.