Re: [PATCH V2] cpufreq: Call transition notifier only once for each policy

[Peter Zijlstra <peterz@infradead.org>]

On Fri, Mar 15, 2019 at 02:43:07PM +0530, Viresh Kumar wrote:
> diff --git a/arch/x86/kernel/tsc.c b/arch/x86/kernel/tsc.c
> index 3fae23834069..cff8779fc0d2 100644
> --- a/arch/x86/kernel/tsc.c
> +++ b/arch/x86/kernel/tsc.c
> @@ -956,28 +956,38 @@ static int time_cpufreq_notifier(struct notifier_block *nb, unsigned long val,
>  				void *data)
>  {
>  	struct cpufreq_freqs *freq = data;
> -	unsigned long *lpj;
> -
> -	lpj = &boot_cpu_data.loops_per_jiffy;
> -#ifdef CONFIG_SMP
> -	if (!(freq->flags & CPUFREQ_CONST_LOOPS))
> -		lpj = &cpu_data(freq->cpu).loops_per_jiffy;
> -#endif
> +	struct cpumask *cpus = freq->policy->cpus;
> +	bool boot_cpu = !IS_ENABLED(CONFIG_SMP) || freq->flags & CPUFREQ_CONST_LOOPS;
> +	unsigned long lpj;
> +	int cpu;
>  
>  	if (!ref_freq) {
>  		ref_freq = freq->old;
> -		loops_per_jiffy_ref = *lpj;
>  		tsc_khz_ref = tsc_khz;
> +
> +		if (boot_cpu)
> +			loops_per_jiffy_ref = boot_cpu_data.loops_per_jiffy;
> +		else
> +			loops_per_jiffy_ref = cpu_data(cpumask_first(cpus)).loops_per_jiffy;
>  	}
> +
>  	if ((val == CPUFREQ_PRECHANGE  && freq->old < freq->new) ||
>  			(val == CPUFREQ_POSTCHANGE && freq->old > freq->new)) {
> -		*lpj = cpufreq_scale(loops_per_jiffy_ref, ref_freq, freq->new);
> -
> +		lpj = cpufreq_scale(loops_per_jiffy_ref, ref_freq, freq->new);
>  		tsc_khz = cpufreq_scale(tsc_khz_ref, ref_freq, freq->new);
> +
>  		if (!(freq->flags & CPUFREQ_CONST_LOOPS))
>  			mark_tsc_unstable("cpufreq changes");
>  
> -		set_cyc2ns_scale(tsc_khz, freq->cpu, rdtsc());
> +		if (boot_cpu) {
> +			boot_cpu_data.loops_per_jiffy = lpj;
> +		} else {
> +			for_each_cpu(cpu, cpus)
> +				cpu_data(cpu).loops_per_jiffy = lpj;
> +		}
> +
> +		for_each_cpu(cpu, cpus)
> +			set_cyc2ns_scale(tsc_khz, cpu, rdtsc());

This code doesn't make sense, the rdtsc() _must_ be called on the CPU in
question. That's part of the whole problem here, TSC isn't sync'ed when
it's subject to CPUFREQ.

>  	}
>  
>  	return 0;