lkml.org 
[lkml]   [2015]   [Sep]   [9]   [last100]   RSS Feed
Views: [wrap][no wrap]   [headers]  [forward] 
 
Messages in this thread
/
From
Date
SubjectRe: [PATCH 3/6] ebpf: add a way to dump an eBPF program
On Wed, Sep 9, 2015 at 5:13 PM, Tycho Andersen
<tycho.andersen@canonical.com> wrote:
> On Wed, Sep 09, 2015 at 04:44:24PM -0700, Andy Lutomirski wrote:
>> On Wed, Sep 9, 2015 at 3:34 PM, Tycho Andersen
>> <tycho.andersen@canonical.com> wrote:
>> >
>> > Here's a thought,
>> >
>> > The set I'm currently proposing effectively separates the ref-counting
>> > of the struct seccomp_filter from the struct bpf_prog (by necessity,
>> > since we're referring to filters from fds). What if we went a little
>> > futher, and made a copy of each seccomp_filter on fork(), keeping it
>> > pointed at the same bpf_prog but adding some metadata about how it was
>> > inherited (tsk->seccomp.filter->inheritence_count++ perhaps). This
>> > would still require this change:
>>
>> Won't that break the tsync mechanism?
>
> We'll need the change I posted (is_ancestor comparing the underlying
> bpf_prog instead of the seccomp_filter), but then I think it'll work.
> I guess we'll need to do some more bookkeeping when we install filters
> via TSYNC since each thread would need its own seccomp_filter, and
> we'd also have to decide whether a filter installed via TSYNC was
> inherited or not.
>
> Am I missing something?

Yes. I don't think that:

int fd = [create an ebpf fd];
if (fork()) {
/* Process A */
seccomp(attach fd);
...
} else {
/* Process B */
seccomp(attach fd);
...
}

should result in processes A and B being considered to have the same
seccomp_filter state. In particular, I eventually want to make the
seccomp_filter state be considerably more interesting than just the
bpf program.

IOW I really do think that seccomp_filter should have identity.

There's another severe problem, I think. Suppose that ebpf1 and ebpf2
are ebpf fds. If processes C and D start out with no filters at all,
C attaches ebpf1 and ebpf2, and D attaches just ebpf2, then C and D
are definitely *not* in the same state, and neither is an ancestor of
the other.

--Andy


\
 
 \ /
  Last update: 2015-09-10 03:01    [W:0.125 / U:0.668 seconds]
©2003-2020 Jasper Spaans|hosted at Digital Ocean and TransIP|Read the blog|Advertise on this site