On Wed, Jun 27, 2018 at 05:10:48AM +0900, Tetsuo Handa wrote: > On 2018/06/27 2:03, Paul E. McKenney wrote: > > There are a lot of ways it could be made concurrency safe. If you need > > me to do this, please do let me know. > > > > That said, the way it is now written, if you invoke rcu_oom_notify() > > twice in a row, the second invocation will wait until the memory from > > the first invocation is freed. What do you want me to do if you invoke > > me concurrently? > > > > 1. One invocation "wins", waits for the earlier callbacks to > > complete, then encourages any subsequent callbacks to be > > processed more quickly. The other invocations return > > immediately without doing anything. > > > > 2. The invocations serialize, with each invocation waiting for > > the callbacks from previous invocation (in mutex_lock() order > > or some such), and then starting a new round. > > > > 3. Something else? > > > > Thanx, Paul > > As far as I can see, > > - atomic_set(&oom_callback_count, 1); > + atomic_inc(&oom_callback_count); > > should be sufficient. I don't see how that helps. For example, suppose that two tasks invoked rcu_oom_notify() at about the same time. Then they could both see oom_callback_count equal to zero, both atomically increment oom_callback_count, then both do the IPI invoking rcu_oom_notify_cpu() on each online CPU. So far, so good. But rcu_oom_notify_cpu() enqueues a per-CPU RCU callback, and enqueuing the same callback twice in quick succession would fatally tangle RCU's callback lists. What am I missing here? Thanx, Paul
