On Wed, Jun 11, 2014 at 07:07:05PM +0200, Oleg Nesterov wrote:
> On 06/11, Paul E. McKenney wrote:
> >
> > @@ -1202,10 +1204,14 @@ static int rcu_boost(struct rcu_node *rnp)
> > t = container_of(tb, struct task_struct, rcu_node_entry);
> > rt_mutex_init_proxy_locked(&mtx, t);
> > t->rcu_boost_mutex = &mtx;
> > + init_completion(&rnp->boost_completion);
>
> can't rcu_init_one() do this? but this is minor,
It could, but I would have to define yet another init-time function under
CONFIG_RCU_BOOST and not. Yeah, lazy...
> > raw_spin_unlock_irqrestore(&rnp->lock, flags);
> > rt_mutex_lock(&mtx); /* Side effect: boosts task t's priority. */
> > rt_mutex_unlock(&mtx); /* Keep lockdep happy. */
> >
> > + /* Wait until boostee is done accessing mtx before reinitializing. */
> > + wait_for_completion(&rnp->boost_completion);
> > +
>
> I must have missed something, I dont understand why we need
> ->boost_completion.
Because rt_mutex_init_proxy_locked() stomps on mtx periodically.
Which might happen to be work at the moment, but doesn't seem like a
particularly good thing.
> What if you simply move that rt_mutex into rcu_node ?
>
> Or. Given that rcu_boost_kthread() never exits, it can declare this mutex
> on stack and pass the pointer to rcu_boost() ?
It is true that moving mtx to either the rcu_node structure or to
rcu_boost_kthread()'s stack frame would preserve type safety, but not
initialization safety.
Or maybe I am being too paranoid?
Thanx, Paul
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