Re: [RFC] kernel/pid.c pid allocation wierdness

Dmitry Adamushko
2007-03-16T11:40:37Z

On 16/03/07, Pavel Emelianov <xemul@sw.ru> wrote:
> Oleg Nesterov wrote:
> > On 03/14, Eric W. Biederman wrote:
> >> Pavel Emelianov <xemul@sw.ru> writes:
> >>
> >>> Hi.
> >>>
> >>> I'm looking at how alloc_pid() works and can't understand
> >>> one (simple/stupid) thing.
> >>>
> >>> It first kmem_cache_alloc()-s a strct pid, then calls
> >>> alloc_pidmap() and at the end it taks a global pidmap_lock()
> >>> to add new pid to hash.
> >
> > We need some global lock. pidmap_lock is already here, and it is
> > only used to protect pidmap->page allocation. Iow, it is almost
> > unused. So it was very natural to re-use it while implementing
> > pidrefs.
> >
> >>> The question is - why does alloc_pidmap() use at least
> >>> two atomic ops and potentially loop to find a zero bit
> >>> in pidmap? Why not call alloc_pidmap() under pidmap_lock
> >>> and find zero pid in pidmap w/o any loops and atomics?
> >
> > Currently we search for zero bit lockless, why do you want
> > to do it under spin_lock ?
>
> Search isn't lockless. Look:
>
> while (1) {
> if (!test_and_set_bit(...)) {
> atomic_dec(&nr_free);
> return pid;
> }
> we use two atomic operations to find and set a bit in a map.

While you may have a few concurrent threads competing for the same
"offset" and "pid" in the loop - e.g. at point [1] (see below), only
one will succeed with "registering" it due to the atomicity of
test_and_set_bit() and so only this one will get at point [2] with the
"pid".

The rest of the "unlucky" threads will either

(i) compete for another "offset" -> "pid" (as described above);
(ii) leave the loop when one of the conditions of while() becomes
"false" -> e.g. there are no more free slots in this map.

if (likely(atomic_read(&map->nr_free))) {
do {
// [1]
if (!test_and_set_bit(offset, map->page)) {
// [2]
atomic_dec(&map->nr_free);
pid_ns->last_pid = pid;
return pid;
}
offset = find_next_offset(map, offset);
pid = mk_pid(pid_ns, map, offset);
/*
* find_next_offset() found a bit, the pid from it
* is in-bounds, and if we fell back to the last
* bitmap block and the final block was the same
* as the starting point, pid is before last_pid.
*/
} while (offset < BITS_PER_PAGE && pid < pid_max &&
(i != max_scan || pid < last ||
!((last+1) & BITS_PER_PAGE_MASK)));
}

> ...

>
> > Oleg.
> >
> >

--
Best regards,
Dmitry Adamushko