Re: [PATCH v1 1/2] perf,x86: add Intel RAPL PMU support

[Stephane Eranian]

Andi,

On Mon, Oct 7, 2013 at 11:45 PM, Andi Kleen <andi@xxxxxxxxxxxxxx> wrote:
> Stephane Eranian <eranian@xxxxxxxxxx> writes:
>>
>>>> +             goto again;
>>>> +
>>>> +     struct rapl_pmu *pmu = __get_cpu_var(rapl_pmu);
>>>> +
>>>> +     if (WARN_ON_ONCE(!(event->hw.state & PERF_HES_STOPPED)))
>>>> +             return;
>>>> +
>>>> +     event->hw.state = 0;
>>>> +
>>>> +     local64_set(&event->hw.prev_count, rapl_read_counter(event));
>>>> +
>>>> +     pmu->n_active++;
>>>
>>> What lock protects this add?
>>>
>> None. I will add one. Bu then I am wondering about if it is really
>> necessary given
>> that RAPL event are system-wide and this pinned to a CPU. If the call came
>> from another CPU, then it IPI there, and that means that CPU is executing that
>> code. Any other CPU will need IPI too, and that interrupt will be kept pending.
>> Am I missing a test case here? Are IPI reentrant?
>
> they can be if interrupts are enabled (likely here)
>
So, I spent some time trying to figure this out via instrumentation and it seems
it is never the case that this function or in fact __perf_event_enable() for a
syswide event is called with interrupts enabled. Why?

Well, it has to do with cpu_function_call() which is ALWAYS called for a syswide
event on the perf_event_enable() code path.

If you are calling for an event on the same CPU, you end up executing:
smp_call_function_single()
       if (cpu == this_cpu) {
                local_irq_save(flags);
                func(info);
                local_irq_restore(flags);

If you are calling a remote CPU, then you end up in the APIC code to send
an IPI. On the receiving side, I could not find the local_irq_save() call, but
I verified that upon entry, __perf_event_enable() has interrupts disabled.
And that's either because I missed the interrupt masking call OR because
the HW does it automatically for us. I could not yet figure this out.

In any case, looks like both the start() and stop() routine are protected
from interrupts and thus preemption, so we may not need a lock to
protect n_active.
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