sched/clock: Remove pointless test in cpu_clock/local_clock

In case the HAVE_UNSTABLE_SCHED_CLOCK config is set, the cpu_clock() version checks if sched_clock_stable() is not set and calls sched_clock_cpu(), otherwise it calls sched_clock().

sched_clock_cpu() checks also if sched_clock_stable() is set and, if true, calls sched_clock().

sched_clock() will be called in sched_clock_cpu() if sched_clock_stable() is true.

Remove the duplicate test by directly calling sched_clock_cpu() and let the static key act in this function instead. We can assume gcc is smart enough to inline cpu_clock()/local_clock() to sched_clock_cpu().

---
 kernel/sched/clock.c | 10 ++--------
 1 file changed, 2 insertions(+), 8 deletions(-)

index fedb967..30c4b20 100644
@@ -375,10 +375,7 @@ EXPORT_SYMBOL_GPL(sched_clock_idle_wakeup_event);
  */
 u64 cpu_clock(int cpu)
 {
-	if (!sched_clock_stable())
-		return sched_clock_cpu(cpu);
-
-	return sched_clock();
+	return sched_clock_cpu(cpu);
 }
 
 /*

@@ -390,10 +387,7 @@ u64 cpu_clock(int cpu)
  */
 u64 local_clock(void)
 {
-	if (!sched_clock_stable())
-		return sched_clock_cpu(raw_smp_processor_id());
-
-	return sched_clock();
+	return sched_clock_cpu(raw_smp_processor_id());
 }
 
 #else /* CONFIG_HAVE_UNSTABLE_SCHED_CLOCK */

1.9.1

Why assume if you can easily check; so does it indeed inline?

Actually, no. They are not inlined.

With this patch we end up with one line functions:

#ifdef CONFIG_HAVE_UNSTABLE_SCHED_CLOCK

u64 cpu_clock(int cpu)

{
	return sched_clock_cpu(cpu);
}

u64 local_clock(void)

{
	return sched_clock_cpu(raw_smp_processor_id());
}

#else

u64 cpu_clock(int cpu)

{
	return sched_clock();
}

u64 local_clock(void)

{
	return sched_clock();
}

#endif

is it worth to convert them to static inline function ?

(static inline does effectively inline those functions).

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