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Date:   Fri, 17 Jul 2020 13:52:46 -0400 (EDT)
From:   Mathieu Desnoyers <mathieu.desnoyers@...icios.com>
To:     Alan Stern <stern@...land.harvard.edu>
Cc:     Nicholas Piggin <npiggin@...il.com>, paulmck <paulmck@...nel.org>,
        Anton Blanchard <anton@...abs.org>,
        Arnd Bergmann <arnd@...db.de>,
        linux-arch <linux-arch@...r.kernel.org>,
        linux-kernel <linux-kernel@...r.kernel.org>,
        linux-mm <linux-mm@...ck.org>,
        linuxppc-dev <linuxppc-dev@...ts.ozlabs.org>,
        Andy Lutomirski <luto@...nel.org>,
        Peter Zijlstra <peterz@...radead.org>, x86 <x86@...nel.org>
Subject: Re: [RFC PATCH 4/7] x86: use exit_lazy_tlb rather than
 membarrier_mm_sync_core_before_usermode

----- On Jul 17, 2020, at 1:44 PM, Alan Stern stern@...land.harvard.edu 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@...land.harvard.edu 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

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