> -----Original Message----- > From: Stephane Eranian [mailto:eran...@google.com] > Sent: Thursday, August 06, 2015 4:38 PM > To: Liang, Kan > Cc: Peter Zijlstra; mi...@redhat.com; Arnaldo Carvalho de Melo; > a...@linux.intel.com; LKML > Subject: Re: [PATCH V2 1/1] perf/x86: Add Intel power cstate PMUs > support > > On Thu, Aug 6, 2015 at 1:25 PM, Liang, Kan <kan.li...@intel.com> wrote: > > > >> >> >> >> +static cpumask_t power_cstate_core_cpu_mask; > >> >> >> > > >> >> >> > That one typically does not need a cpumask. > >> >> >> > > >> >> >> You need to pick one CPU out of the multi-core. But it is for > >> >> >> client parts thus there is only one socket. At least this is my > >> understanding. > >> >> >> > >> >> > > >> >> > CORE_C*_RESIDENCY are available for physical processor core. > >> >> > So logical processor in same physical processor core share the > >> >> > same counter. > >> >> > I think we need the cpumask to identify the default logical > >> >> > processor which do counting. > >> >> > > >> >> Did you restrict these events to system-wide mode only? > >> >> > >> Ok, so that means that your cpumask includes one HT per physical core. > >> But then, the result is not the simple aggregation of all the N/2 CPUs. > > > > The counter counts per physical core. The result is the aggregation of > > all HT cpus in same physical core. > > But then don't you need to divide by 2 to get a meaningful result?
Rethink of it. I think I was unclear about the aggregation of all HT cpus in same physical core. physical core Cstate should equal to min(logical core C-state). So only all logical core enters C6-state, the physical core enters C6-state, then CORE_C6_RESIDENCY counts. So if we only count on one logical core/HT for CORE_C6_RESIDENCY. We don't need to divide by 2. The count result is the residency when all logical core in C6 (some may deeper). Thanks, Kan
