Re: NTFS umask option problem

Martin v. Loewis (loewis@informatik.hu-berlin.de)
Wed, 28 Jan 1998 14:10:36 +0100


Hi Dave,

Sorry, but I fail to see the problem. From mount(8)

Set the umask (the bitmask of the permissions that
are not present). The default is the umask of the
current process. The value is given in octal.

The umask of the current process is not considered, but I believe it
is not for any other file system supporting the umask mount option.

> [root@gatsby davew]# mount -t ntfs -o umask=0444 /dev/hda1 /mnt/nt
> [root@gatsby davew]# ls -l /mnt/nt
> total 20
> d--x--x--x 1 root root 4096 Mar 9 1997 Program Files
> d--x--x--x 1 root root 20480 Dec 15 21:13 archive
> ---------- 1 root root 0 Nov 14 1996 autoexec.bat

So you request that the read bits (4) are not present. And alas, there
are no read bits set.

> [root@gatsby davew]# umount /mnt/nt
> [root@gatsby davew]# mount -t ntfs -o umask=0222 /dev/hda1 /mnt/nt
> [root@gatsby davew]# ls -l /mnt/nt
> total 16
> dr-xr-xr-x 1 root root 4096 Mar 9 1997 Program Files
> dr-xr-xr-x 1 root root 20480 Dec 15 21:13 archive
> -r--r--r-- 1 root root 0 Nov 14 1996 autoexec.bat
> -r--r--r-- 1 root root 278 Nov 18 11:52 boot.ini

Here, you request that the write bits are not set. And see, there are
no write bits anywhere.

The confusing thing may be that there will be *never* write bits,
unless you enabled the experimental write support (and run a version
of the driver that will set the write bits in this case).

If this is not an answer to your question, please rephrase it.

Martin