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Message-ID: <1d03d443-b187-bc1f-2601-a54037a64eff@linux.intel.com>
Date: Fri, 7 Aug 2020 14:24:30 +0800
From: "Jin, Yao" <yao.jin@...ux.intel.com>
To: peterz@...radead.org
Cc: mingo@...hat.com, oleg@...hat.com, acme@...nel.org,
jolsa@...nel.org, Linux-kernel@...r.kernel.org, ak@...ux.intel.com,
kan.liang@...el.com, yao.jin@...el.com,
alexander.shishkin@...ux.intel.com, mark.rutland@....com
Subject: Re: [PATCH v1 2/2] perf/core: Fake regs for leaked kernel samples
Hi Peter,
On 8/6/2020 7:00 PM, peterz@...radead.org wrote:
> On Thu, Aug 06, 2020 at 11:18:27AM +0200, peterz@...radead.org wrote:
>
>> Suppose we have nested virt:
>>
>> L0-hv
>> |
>> G0/L1-hv
>> |
>> G1
>>
>> And we're running in G0, then:
>>
>> - 'exclude_hv' would exclude L0 events
>> - 'exclude_host' would ... exclude L1-hv events?
>> - 'exclude_guest' would ... exclude G1 events?
>
> So in arch/x86/events/intel/core.c we have:
>
> static inline void intel_set_masks(struct perf_event *event, int idx)
> {
> struct cpu_hw_events *cpuc = this_cpu_ptr(&cpu_hw_events);
>
> if (event->attr.exclude_host)
> __set_bit(idx, (unsigned long *)&cpuc->intel_ctrl_guest_mask);
> if (event->attr.exclude_guest)
> __set_bit(idx, (unsigned long *)&cpuc->intel_ctrl_host_mask);
> if (event_is_checkpointed(event))
> __set_bit(idx, (unsigned long *)&cpuc->intel_cp_status);
> }
>
exclude_host is now set by guest (pmc_reprogram_counter, arch/x86/kvm/pmu.c). When enabling the
event, we can check exclude_host to know if it's a guest.
Otherwise we may need more flags in event->attr to indicate the status.
> which is, afaict, just plain wrong. Should that not be something like:
>
> if (!event->attr.exclude_host)
> __set_bit(idx, (unsigned long *)&cpuc->intel_ctrl_host_mask);
> if (!event->attr.exclude_guest)
> __set_bit(idx, (unsigned long *)&cpuc->intel_ctrl_guest_mask);
>
>
How can we know it's guest or host even if exclude_host is set in guest?
Thanks
Jin Yao
> Also, ARM64 seems to also implement this stuff, Mark, do you have any
> insight on how all this is 'supposed' to work?
>
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