----- On Jul 17, 2020, at 1:44 PM, Alan Stern stern@xxxxxxxxxxxxxxxxxxx wrote: > On Fri, Jul 17, 2020 at 12:22:49PM -0400, Mathieu Desnoyers wrote: >> ----- On Jul 17, 2020, at 12:11 PM, Alan Stern stern@xxxxxxxxxxxxxxxxxxx wrote: >> >> >> > I agree with Nick: A memory barrier is needed somewhere between the >> >> > assignment at 6 and the return to user mode at 8. Otherwise you end up >> >> > with the Store Buffer pattern having a memory barrier on only one side, >> >> > and it is well known that this arrangement does not guarantee any >> >> > ordering. >> >> >> >> Yes, I see this now. I'm still trying to wrap my head around why the memory >> >> barrier at the end of membarrier() needs to be paired with a scheduler >> >> barrier though. >> > >> > The memory barrier at the end of membarrier() on CPU0 is necessary in >> > order to enforce the guarantee that any writes occurring on CPU1 before >> > the membarrier() is executed will be visible to any code executing on >> > CPU0 after the membarrier(). Ignoring the kthread issue, we can have: >> > >> > CPU0 CPU1 >> > x = 1 >> > barrier() >> > y = 1 >> > r2 = y >> > membarrier(): >> > a: smp_mb() >> > b: send IPI IPI-induced mb >> > c: smp_mb() >> > r1 = x >> > >> > The writes to x and y are unordered by the hardware, so it's possible to >> > have r2 = 1 even though the write to x doesn't execute until b. If the >> > memory barrier at c is omitted then "r1 = x" can be reordered before b >> > (although not before a), so we get r1 = 0. This violates the guarantee >> > that membarrier() is supposed to provide. >> > >> > The timing of the memory barrier at c has to ensure that it executes >> > after the IPI-induced memory barrier on CPU1. If it happened before >> > then we could still end up with r1 = 0. That's why the pairing matters. >> > >> > I hope this helps your head get properly wrapped. :-) >> >> It does help a bit! ;-) >> >> This explains this part of the comment near the smp_mb at the end of membarrier: >> >> * Memory barrier on the caller thread _after_ we finished >> * waiting for the last IPI. [...] >> >> However, it does not explain why it needs to be paired with a barrier in the >> scheduler, clearly for the case where the IPI is skipped. I wonder whether this >> part >> of the comment is factually correct: >> >> * [...] Matches memory barriers around rq->curr modification in scheduler. > > The reasoning is pretty much the same as above: > > CPU0 CPU1 > x = 1 > barrier() > y = 1 > r2 = y > membarrier(): > a: smp_mb() > switch to kthread (includes mb) > b: read rq->curr == kthread > switch to user (includes mb) > c: smp_mb() > r1 = x > > Once again, it is possible that x = 1 doesn't become visible to CPU0 > until shortly before b. But if c is omitted then "r1 = x" can be > reordered before b (to any time after a), so we can have r1 = 0. > > Here the timing requirement is that c executes after the first memory > barrier on CPU1 -- which is one of the ones around the rq->curr > modification. (In fact, in this scenario CPU1's switch back to the user > process is irrelevant.) That indeed covers the last scenario I was wondering about. Thanks Alan! Mathieu > > Alan Stern -- Mathieu Desnoyers EfficiOS Inc. http://www.efficios.com
