Messages in this thread | | | Date | Mon, 26 Sep 2016 11:35:03 -0400 | From | Steven Rostedt <> | Subject | Re: [PATCH -v2 1/9] rtmutex: Deboost before waking up the top waiter |
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On Mon, 26 Sep 2016 17:22:28 +0200 Peter Zijlstra <peterz@infradead.org> wrote:
> > > + /* > > > + * We should deboost before waking the top waiter task such that > > > + * we don't run two tasks with the 'same' priority. This however > > > + * can lead to prio-inversion if we would get preempted after > > > + * the deboost but before waking our high-prio task, hence the > > > + * preempt_disable before unlock. Pairs with preempt_enable() in > > > + * rt_mutex_postunlock(); > > > > There's a preempt_enable() in rt_mutex_postunlock()? Does > > wake_futex_pi() know that? > > > > Not sure I see your point. rt_mutex_futex_unlock() calls > rt_mutex_slowunlock() which does the preempt_disable(), we then pass the > return of that into deboost, which we pass into rt_mutex_postunlock() > and everything should be balanced.
Can we please add more comments explaining this. Having side effects of functions disabling preemption, passing a bool saying that it did, and needing to call another function (somewhat seemingly unrelated) to re-enable preemption, just seems a bit of a stretch for maintainable code.
Especially now that the code after the spin_unlock(&hb->lock) is now a critical section (preemption is disable). There's nothing obvious in futex.c that says it is.
Just think about looking at this code in another 5 years. Are you going to remember all this?
-- Steve
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