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SubjectRE: [PATCH v2 6/8] perf/x86/intel/lbr: guest requesting KVM for lbr stack save/restore
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On Saturday, September 8, 2018 4:05 AM, Andi Kleen wrote:
> > How would you realize the function of saving/restoring the lbr stack on the
> host?
> >
> > Here, we create a perf event on the host (please see
> guest_lbr_event_create on patch 7), which essentially satisfies all the
> conditions (e.g. increases cpuc->lbr_users) that are required to have the lbr
> stack saved/restored on the vCPU switching.
> >
> > If we want to stop the host side lbr stack save/restore for the vCPU, we
> need accordingly to call guest_lbr_event_release (in patch 7) to destroy that
> perf event (the host doesn't automatically stop saving the lbr stack for the
> vCPU if that perf event is still there).
> >
> > When would you call that release function? (we all know that the lbr
> > doesn't need to be saved when the guest is not using it, but we need
> > to destroy that perf event to achieve "doesn't need to be saved")
>
> Maybe set a timer on DEBUGCTL LBR=0 ? A timer would provide hysteresis,
> so that quick toggles (like in a PMI handler) wouldn't do anything expensive.

I'm not sure if this would solve the key problem. What would be the frequency to have the timer fired?

- Let's say every 10ms, for example. The guest is disturbed by a timer interrupt (cause VMExit) every 10ms, though the guest doesn't use the lbr any more after the first use. The problem is switched to when do we call the release function to cancel the timer if we want to avoid that unnecessary disturbance to the guest.

- When the timer fires, and it finds "DEBUGCTL LBR=0", it destroys the host side perf event, then the lbr stack isn't saved when the vCPU is scheduled out. As also mentioned in the commit log, perf_pmu_sched_task in the guest disables that bit before reading out the lbr stack (pmi is another example). Now, DEBUGCTL LBR=0", and before the guest read out the lbr stack, the vCPU may happen to be scheduled out, and another thread on the host is scheduled in and will get the lbr stack overwritten. So, before the guest reads out the lbr stack, the stack has already been polluted in this case.

Best,
Wei

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