Re: [PATCH v4 6/7] cpufreq: Support for fast frequency switching

Rafael J. Wysocki
2016-03-16T16:58:22Z

On Wed, Mar 16, 2016 at 4:43 PM, Peter Zijlstra <peterz@infradead.org> wrote:
> On Wed, Mar 16, 2016 at 03:52:28PM +0100, Rafael J. Wysocki wrote:
>> +void cpufreq_enable_fast_switch(struct cpufreq_policy *policy)
>> +{
>> + mutex_lock(&cpufreq_fast_switch_lock);
>> + if (policy->fast_switch_possible && cpufreq_fast_switch_count >= 0) {
>> + cpufreq_fast_switch_count++;
>> + policy->fast_switch_enabled = true;
>> + } else {
>> + pr_warn("cpufreq: CPU%u: Fast freqnency switching not enabled\n",
>> + policy->cpu);
>
> This happens because there's transition notifiers, right? Would it make
> sense to iterate the notifier here and print the notifier function
> symbol for each? That way we've got a clue as to where to start looking
> when this happens.

OK

>> + }
>> + mutex_unlock(&cpufreq_fast_switch_lock);
>> +}
>
>> @@ -1653,8 +1703,18 @@ int cpufreq_register_notifier(struct not
>>
>> switch (list) {
>> case CPUFREQ_TRANSITION_NOTIFIER:
>> + mutex_lock(&cpufreq_fast_switch_lock);
>> +
>> + if (cpufreq_fast_switch_count > 0) {
>> + mutex_unlock(&cpufreq_fast_switch_lock);
>
> So while theoretically (it has a return code)
> cpufreq_register_notifier() could fail, it never actually did. Now we
> do. Do we want to add a WARN here?

Like if (WARN_ON(cpufreq_fast_switch_count > 0)) {

That can be done. :-)