Re: [RFC PATCH 5/6] bpf: Implement function bpf_read_pmu() that get the selected hardware PMU conuter

Wangnan (F)
2015-07-17T11:30:46Z

On 2015/7/17 19:05, Peter Zijlstra wrote:
> On Fri, Jul 17, 2015 at 06:43:35PM +0800, kaixu xia wrote:
>> The function bpf_read_pmu() can get the specific map key, convert
>> the corresponding map value to the pointer to struct perf_event and
>> return the Hardware PMU counter value.
> Thanks for having me on Cc :/
>
>> Signed-off-by: kaixu xia <xiakaixu@huawei.com>
>> ---
>> +static u64 bpf_read_pmu(u64 r1, u64 r2, u64 r3, u64 r4, u64 r5)
>> +{
>> + void *value = (void *) (unsigned long) r1;
>> + struct perf_event *event;
>> + u64 count;
>> +
>> + if (!value || !(*(unsigned long *)value))
>> + return 0;
>> +
>> + event = (struct perf_event *)(*(unsigned long *)value);
>> +
>> + if (event->state == PERF_EVENT_STATE_ACTIVE)
>> + event->pmu->read(event);
>> +
>> + count = local64_read(&event->count);
>> +
>> + return count;
>> +}
> Hell no, that's way broken.
What about calling perf_event_read_value() then?

...
struct perf_event_context *ctx;
u64 enabled, u64 running

ctx = perf_event_ctx_lock(event);
if (!event->state == PERF_EVENT_STATE_ERROR) {
count = perf_event_read_value(event, &enable, &running);
}
perf_event_ctx_unlock(event, ctx);
...

Code is from perf_read().

Thank you.