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    SubjectRe: [patch 1/8] add user mounts to the kernel
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    > >> > +	if (mnt->mnt_flags & MNT_USER)
    > >> > + seq_printf(m, ",user=%i", mnt->mnt_uid);
    > >> How about making the test "if (mnt->mnt_user != &root_user)"
    > >
    > > We don't want to treat root_user special. That's what capabilities
    > > were invented for.
    >
    > For the print statement? What ever it is minor.

    It is a user interface, not a print statement. Your suggested change
    would be vetoed by any number of people.

    So either we have all mounts having owners, AND have /proc/mounts add
    "user=0" to all mounts. While I don't _think_ this would actually
    break userspace, it would definitely make people complain.

    The other choice is what the current patchset does: is to have
    "legacy" mounts without owners, and "new generation" mounts with
    owners having "user=UID" in /proc/mounts, regardless of the value of
    UID.

    > So I want to minimize the changes needed to existing programs.
    > Now if all we have to do is specify MS_SETUSER when root a
    > user with CAP_SETUID is setting up a mount as a user other
    > then himself then I don't much care. If we have to call MS_SETUSER
    > as unprivileged users

    You don't. Unprivileged mounts _imply_ MS_SETUSER.

    > >> > +
    > >> > + uid_t mnt_uid; /* owner of the mount */
    > >>
    > >> Can we please make this a user struct. That requires a bit of
    > >> reference counting but it has uid namespace benefits as well
    > >> as making it easy to implement per user mount rlimits.
    > >
    > > OK, can you ellaborate, what the uid namespace benifits are?
    >
    > In the uid namespace the comparison is simpler as are the propagations
    > rules. Basically if you use a struct user you will never need to
    > care about a uid namespace. If you don't we will have to tear through
    > this code another time.

    Well, OK. I'll do the user_struct thing then.

    Miklos
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