Re: [PATCH 5/6] kmod: preempt on kmod_umh_threads_get()

From: Luis R. Rodriguez
Date: Wed May 24 2017 - 20:15:13 EST


On Fri, May 19, 2017 at 03:27:12PM -0700, Dmitry Torokhov wrote:
> On Thu, May 18, 2017 at 08:24:43PM -0700, Luis R. Rodriguez wrote:
> > In theory it is possible multiple concurrent threads will try to
> > kmod_umh_threads_get() and as such atomic_inc(&kmod_concurrent) at
> > the same time, therefore enabling a small time during which we've
> > bumped kmod_concurrent but have not really enabled work. By using
> > preemption we mitigate this a bit.
> >
> > Preemption is not needed when we kmod_umh_threads_put().
> >
> > Signed-off-by: Luis R. Rodriguez <mcgrof@xxxxxxxxxx>
> > ---
> > kernel/kmod.c | 24 ++++++++++++++++++++++--
> > 1 file changed, 22 insertions(+), 2 deletions(-)
> >
> > diff --git a/kernel/kmod.c b/kernel/kmod.c
> > index 563600fc9bb1..7ea11dbc7564 100644
> > --- a/kernel/kmod.c
> > +++ b/kernel/kmod.c
> > @@ -113,15 +113,35 @@ static int call_modprobe(char *module_name, int wait)
> >
> > static int kmod_umh_threads_get(void)
> > {
> > + int ret = 0;
> > +
> > + /*
> > + * Disabling preemption makes sure that we are not rescheduled here
> > + *
> > + * Also preemption helps kmod_concurrent is not increased by mistake
> > + * for too long given in theory two concurrent threads could race on
> > + * atomic_inc() before we atomic_read() -- we know that's possible
> > + * and but we don't care, this is not used for object accounting and
> > + * is just a subjective threshold. The alternative is a lock.
> > + */
> > + preempt_disable();
> > atomic_inc(&kmod_concurrent);
> > if (atomic_read(&kmod_concurrent) <= max_modprobes)
>
> That is very "fancy" way to basically say:
>
> if (atomic_inc_return(&kmod_concurrent) <= max_modprobes)

Do you mean to combine the atomic_inc() and atomic_read() in one as you noted
(as that is not a change in this patch), *or* that using a memory barrier here
with atomic_inc_return() should suffice to address the same and avoid an
explicit preemption enable / disable ?

Luis