Re: Mount namespace "dominant peer group"?

From: Michael Kerrisk (man-pages)
Date: Sat May 21 2016 - 08:50:08 EST


Hello Ram,

On 05/20/2016 06:15 PM, Ram Pai wrote:
> On Fri, May 20, 2016 at 04:24:18PM -0500, Michael Kerrisk (man-pages) wrote:
>> Hello Miklos,
>>
>> I'm working on some better documentation of mount namespaces,
>> and there's a detail that puzzles me, and I hope you might be
>> able to help, since you added the detail...
>>
>> In Documentation/filesystems/proc.txt there is this text in the
>> description of /proc/PID/mountinfo:
>>
>> [[
>> Parsers should ignore all unrecognised optional fields. Currently the
>> possible optional fields are:
>>
>> shared:X mount is shared in peer group X
>> master:X mount is slave to peer group X
>> propagate_from:X mount is slave and receives propagation from peer group X (*)
>> unbindable mount is unbindable
>>
>> (*) X is the closest dominant peer group under the process's root. If
>> X is the immediate master of the mount, or if there's no dominant peer
>> group under the same root, then only the "master:X" field is present
>> and not the "propagate_from:X" field.
>> ]]
>>
>> What is a dominant peer group, as distinct from the immediate master?
>>
>> I can see in fs/proc_namespaces.c that there is this distinction made:
>>
>> [[
>> /* Tagged fields ("foo:X" or "bar") */
>> if (IS_MNT_SHARED(r))
>> seq_printf(m, " shared:%i", r->mnt_group_id);
>> if (IS_MNT_SLAVE(r)) {
>> int master = r->mnt_master->mnt_group_id;
>> int dom = get_dominating_id(r, &p->root);
>> seq_printf(m, " master:%i", master);
>> if (dom && dom != master)
>> seq_printf(m, " propagate_from:%i", dom);
>> }
>> ]]
>>
>> But I can't relate that to some user-space semantics. I suppose another
>> way of asking my question is: how could I create a slave that is
>> propagating from a peer group other than it's immediate master?
>
> It can happen if you have unmounted or privatised all your master mounts from the peer group.
>
> Eg:
>
> mount /dev/xyz /1 #creates a new mount
> mount --make-private /1 #just make sure that it does not receive or send and propogation
> mount --make-shared /1 #now make it shared.
> mount --bind /1 /2 #create a peer /1 and /2 are peers
> create a new fs-namespace. this new fs-namespace which will have /1' and /2'. /1 /2 /1' /2' are now all part of the same peergroup.
> mount --make-slave /2 # this will make /2 a slave of the peer group that contains /1 /1' and /2'
> umount /1 # we now have /2 which receives propagation from a peer group which does not have a representative in its fs-namespace.

Thanks for the note. However, doing the above, I still do not
see any mount being marked with 'propagate_from'. Perhaps I
misunderstood your instructions above. Here's what I did:

sh1# mount --make-private / # Make share everything is private...
sh1# mount /dev/sdb6 /1
sh1# mount --make-private /1
sh1# mount --make-shared /1
sh1# mount --bind /1 /2
sh1# cat /proc/self/mountinfo | grep '/[12] ' | sed 's/ - .*//'
81 61 8:22 / /1 rw,relatime shared:1
82 61 8:22 / /2 rw,relatime shared:1

Then, at a second terminal, create a new mount NS:

sh2# unshare -m --propagation unchanged sh
sh2# cat /proc/self/mountinfo | grep '/[12] ' | sed 's/ - .*//'
169 132 8:22 / /1 rw,relatime shared:1
170 132 8:22 / /2 rw,relatime shared:1

Returning to the first terminal:

sh1# mount --make-slave /2
sh1# umount /1
sh1# cat /proc/self/mountinfo | grep '/[12] ' | sed 's/ - .*//'
82 61 8:22 / /2 rw,relatime master:1

That is, we see /2 in the initial mount namespace is a slave
but there is no 'propagate_from' tag. Did I miss something?

Cheers,

Michael

--
Michael Kerrisk
Linux man-pages maintainer; http://www.kernel.org/doc/man-pages/
Linux/UNIX System Programming Training: http://man7.org/training/