On Tue, 28 Jan 2025 22:08:27 +0000,
Colton Lewis <coltonlewis@xxxxxxxxxx> wrote:
>
> >> + bitmap_set(cpu_pmu->cntr_mask, 0, pmcr_n);
> >> +
> >> + if (reserved_guest_counters > 0 && reserved_guest_counters < pmcr_n) {
> >> + cpu_pmu->hpmn = reserved_guest_counters;
> >> + cpu_pmu->partitioned = true;
>
> > Isn't this going to completely explode on a kernel running at EL1?
>
> Trying to access an EL2 register at EL1 can do that. I'll add the
> appropriate hypercalls.
But what about a guest that would get passed the magic parameter? I
think you want to prevent that too.
>
> > Also, how does it work in an asymmetric configuration where some CPUs
> > can satisfy the reservation, and some can't?
>
> The CPUs that can't read their own value of PMCR.N below what the
> attempted reservation is and so do not get partitioned. Nothing changes
> for that CPU if it can't meet the reservation.
That's not what I meant. The question really is:
- do we want the reservation to be the number of counters reserved for
the host
- or do we want it to be for the guest?
On symmetric systems, it doesn't matter. On broken big-little systems,
this changes everything (it has a direct impact on userspace's ability
to use the counters).
I think the design should reflect a decision on the above.
M.
--
Without deviation from the norm, progress is not possible.
