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

[Alan Stern]

On Thu, Jul 16, 2020 at 02:58:41PM -0400, Mathieu Desnoyers wrote:
> ----- On Jul 16, 2020, at 12:03 PM, Mathieu Desnoyers mathieu.desnoyers at efficios.com wrote:
> 
> > ----- On Jul 16, 2020, at 11:46 AM, Mathieu Desnoyers
> > mathieu.desnoyers at efficios.com wrote:
> > 
> >> ----- On Jul 16, 2020, at 12:42 AM, Nicholas Piggin npiggin at gmail.com wrote:
> >>> 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.

Alan Stern