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

[Alan Stern]

On Fri, Jul 17, 2020 at 09:39:25AM -0400, Mathieu Desnoyers wrote:
> ----- On Jul 16, 2020, at 5:24 PM, Alan Stern stern at rowland.harvard.edu wrote:
> 
> > 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

...

> >> 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.

...

> > 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.

The key part of my reply was the question: "Is it correct to say that 
the switch_to operations in 5 and 7 include memory barriers?"  From the 
text quoted above and from Nick's reply, it seems clear that they do 
not.

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.

One thing I don't understand about all this: Any context switch has to 
include a memory barrier somewhere, but both you and Nick seem to be 
saying that steps 6 and 7 don't include (or don't need) any memory 
barriers.  What am I missing?

Alan Stern