Thread (59 messages) 59 messages, 6 authors, 2020-07-21

Re: [RFC PATCH 4/7] x86: use exit_lazy_tlb rather than membarrier_mm_sync_core_before_usermode

From: Mathieu Desnoyers <mathieu.desnoyers@efficios.com>
Date: 2020-07-17 13:39:35
Also in: linux-arch, linux-mm, lkml

----- On Jul 16, 2020, at 5:24 PM, Alan Stern stern@rowland.harvard.edu wrote:
On Thu, Jul 16, 2020 at 02:58:41PM -0400, Mathieu Desnoyers wrote:
quoted
----- On Jul 16, 2020, at 12:03 PM, Mathieu Desnoyers
mathieu.desnoyers@efficios.com wrote:
quoted
----- On Jul 16, 2020, at 11:46 AM, Mathieu Desnoyers
mathieu.desnoyers@efficios.com wrote:
quoted
----- On Jul 16, 2020, at 12:42 AM, Nicholas Piggin npiggin@gmail.com wrote:
quoted
I should be more complete here, especially since I was complaining
about unclear barrier comment :)


CPU0                     CPU1
a. user stuff            1. user stuff
b. membarrier()          2. enter kernel
c. smp_mb()              3. smp_mb__after_spinlock(); // in __schedule
d. read rq->curr         4. rq->curr switched to kthread
e. is kthread, skip IPI  5. switch_to kthread
f. return to user        6. rq->curr switched to user thread
g. user stuff            7. switch_to user thread
                        8. exit kernel
                        9. more user stuff

What you're really ordering is a, g vs 1, 9 right?

In other words, 9 must see a if it sees g, g must see 1 if it saw 9,
etc.

Userspace does not care where the barriers are exactly or what kernel
memory accesses might be being ordered by them, so long as there is a
mb somewhere between a and g, and 1 and 9. Right?
This is correct.
Actually, sorry, the above is not quite right. It's been a while
since I looked into the details of membarrier.

The smp_mb() at the beginning of membarrier() needs to be paired with a
smp_mb() _after_ rq->curr is switched back to the user thread, so the
memory barrier is between store to rq->curr and following user-space
accesses.

The smp_mb() at the end of membarrier() needs to be paired with the
smp_mb__after_spinlock() at the beginning of schedule, which is
between accesses to userspace memory and switching rq->curr to kthread.

As to *why* this ordering is needed, I'd have to dig through additional
scenarios from https://lwn.net/Articles/573436/. Or maybe Paul remembers ?
Thinking further about this, I'm beginning to consider that maybe we have been
overly cautious by requiring memory barriers before and after store to rq->curr.

If CPU0 observes a CPU1's rq->curr->mm which differs from its own process
(current)
while running the membarrier system call, it necessarily means that CPU1 had
to issue smp_mb__after_spinlock when entering the scheduler, between any
user-space
loads/stores and update of rq->curr.

Requiring a memory barrier between update of rq->curr (back to current process's
thread) and following user-space memory accesses does not seem to guarantee
anything more than what the initial barrier at the beginning of __schedule
already
provides, because the guarantees are only about accesses to user-space memory.

Therefore, with the memory barrier at the beginning of __schedule, just
observing that
CPU1's rq->curr differs from current should guarantee that a memory barrier was
issued
between any sequentially consistent instructions belonging to the current
process on
CPU1.

Or am I missing/misremembering an important point here ?
Is it correct to say that the switch_to operations in 5 and 7 include
memory barriers?  If they do, then skipping the IPI should be okay.

The reason is as follows: The guarantee you need to enforce is that
anything written by CPU0 before the membarrier() will be visible to CPU1
after it returns to user mode.  Let's say that a writes to X and 9
reads from X.

Then we have an instance of the Store Buffer pattern:

CPU0			CPU1
a. Write X		6. Write rq->curr for user thread
c. smp_mb()		7. switch_to memory barrier
d. Read rq->curr	9. Read X

In this pattern, the memory barriers make it impossible for both reads
to miss their corresponding writes.  Since d does fail to read 6 (it
sees the earlier value stored by 4), 9 must read a.

The other guarantee you need is that g on CPU0 will observe anything
written by CPU1 in 1.  This is easier to see, using the fact that 3 is a
memory barrier and d reads from 4.
Right, and Nick's reply involving pairs of loads/stores on each side
clarifies the situation even further.

Thanks,

Mathieu


-- 
Mathieu Desnoyers
EfficiOS Inc.
http://www.efficios.com
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