On Thu, Oct 05, 2017 at 09:17:03AM -0400, Steven Rostedt wrote: > On Wed, 4 Oct 2017 14:29:27 -0700 > "Paul E. McKenney" <[email protected]> wrote: > > > Consider the following admittedly improbable sequence of events: > > > > o RCU is initially idle. > > > > o Task A on CPU 0 executes rcu_read_lock(). > > A starts rcu_read_lock() critical section. > > > > > o Task B on CPU 1 executes synchronize_rcu(), which must > > wait on Task A: > > B waits for A. > > > > > o Task B registers the callback, which starts a new > > grace period, awakening the grace-period kthread > > on CPU 3, which immediately starts a new grace period. > > [ isn't B blocked (off rq)? How does it migrate? ]
No, its running synchronize_rcu() but hasn't blocked yet. It would block on wait_for_completion(), but per the very last point, we'll observe the complete() before we block. > > o Task B migrates to CPU 2, which provides a quiescent > > state for both CPUs 1 and 2. > > > > o Both CPUs 1 and 2 take scheduling-clock interrupts, > > and both invoke RCU_SOFTIRQ, both thus learning of the > > new grace period. > > > > o Task B is delayed, perhaps by vCPU preemption on CPU 2. > > > > o CPUs 2 and 3 pass through quiescent states, which are reported > > to core RCU. > > > > o Task B is resumed just long enough to be migrated to CPU 3, > > and then is once again delayed. > > > > o Task A executes rcu_read_unlock(), exiting its RCU read-side > > critical section. > > A calls rcu_read_unlock() ending the critical section The point is that rcu_read_unlock() doesn't have memory ordering. > > > > o CPU 0 passes through a quiescent sate, which is reported to > > core RCU. Only CPU 1 continues to block the grace period. > > > > o CPU 1 passes through a quiescent state, which is reported to > > core RCU. This ends the grace period, and CPU 1 therefore > > invokes its callbacks, one of which awakens Task B via > > complete(). > > > > o Task B resumes (still on CPU 3) and starts executing > > wait_for_completion(), which sees that the completion has > > already completed, and thus does not block. It returns from > > the synchronize_rcu() without any ordering against the > > end of Task A's RCU read-side critical section. > > B runs > > > > > > It can therefore mess up Task A's RCU read-side critical section, > > in theory, anyway. > > I don't see how B ran during A's critical section. It didn't but that doesn't mean the memory ordering agrees. What we require is B observes (per the memory ordering) everything up to and including the rcu_read_unlock(). This is not 'time' related. That said, I don't think it can actually happen, because CPU0's QS state is ordered against the complete and the wait_for_completion is ordered against the complete.
