Messages in this thread | | | Subject | Re: [PATCH V5 16/25] perf/x86: Register hybrid PMUs | From | "Liang, Kan" <> | Date | Fri, 9 Apr 2021 09:50:20 -0400 |
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On 4/9/2021 2:58 AM, Peter Zijlstra wrote: > On Mon, Apr 05, 2021 at 08:10:58AM -0700, kan.liang@linux.intel.com wrote: >> @@ -2089,9 +2119,46 @@ static int __init init_hw_perf_events(void) >> if (err) >> goto out1; >> >> - err = perf_pmu_register(&pmu, "cpu", PERF_TYPE_RAW); >> - if (err) >> - goto out2; >> + if (!is_hybrid()) { >> + err = perf_pmu_register(&pmu, "cpu", PERF_TYPE_RAW); >> + if (err) >> + goto out2; >> + } else { >> + u8 cpu_type = get_this_hybrid_cpu_type(); >> + struct x86_hybrid_pmu *hybrid_pmu; >> + bool registered = false; >> + int i; >> + >> + if (!cpu_type && x86_pmu.get_hybrid_cpu_type) >> + cpu_type = x86_pmu.get_hybrid_cpu_type(); >> + >> + for (i = 0; i < x86_pmu.num_hybrid_pmus; i++) { >> + hybrid_pmu = &x86_pmu.hybrid_pmu[i]; >> + >> + hybrid_pmu->pmu = pmu; >> + hybrid_pmu->pmu.type = -1; >> + hybrid_pmu->pmu.attr_update = x86_pmu.attr_update; >> + hybrid_pmu->pmu.capabilities |= PERF_PMU_CAP_HETEROGENEOUS_CPUS; >> + >> + err = perf_pmu_register(&hybrid_pmu->pmu, hybrid_pmu->name, >> + (hybrid_pmu->cpu_type == hybrid_big) ? PERF_TYPE_RAW : -1); >> + if (err) >> + continue; >> + >> + if (cpu_type == hybrid_pmu->cpu_type) >> + x86_pmu_update_cpu_context(&hybrid_pmu->pmu, raw_smp_processor_id()); >> + >> + registered = true; >> + } >> + >> + if (!registered) { >> + pr_warn("Failed to register hybrid PMUs\n"); >> + kfree(x86_pmu.hybrid_pmu); >> + x86_pmu.hybrid_pmu = NULL; >> + x86_pmu.num_hybrid_pmus = 0; >> + goto out2; >> + } > > I don't think this is quite right. registered will be true even if one > fails, while I think you meant to only have it true when all (both) > types registered correctly.
No, I mean that perf error out only when all types fail to be registered.
For the case (1 failure, 1 success), users can still access the registered PMU. When a CPU belongs to the unregistered PMU online, a warning will be displayed. Because in init_hybrid_pmu(), we will check the PMU type before update the CPU mask.
if (WARN_ON_ONCE(!pmu || (pmu->pmu.type == -1))) { cpuc->pmu = NULL; return false; }
Thanks, Kan
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