sched: Avoid printing sched_group::__cpu_power for default case

Commit 46e0bb9c12 ("sched: Print sched_group::__cpu_power
in sched_domain_debug") produces a messy dmesg output while
attempting to print the sched_group::__cpu_power for each
group in the sched_domain hierarchy.

Fix this by avoid printing the __cpu_power for default cases.
(i.e, __cpu_power == SCHED_LOAD_SCALE).

[ Impact: reduce syslog clutter ]

Reported-by: Tony Luck <tony.luck@intel.com>
Signed-off-by: Gautham R Shenoy <ego@in.ibm.com>
Fixed-by: Tony Luck <tony.luck@intel.com>
Cc: a.p.zijlstra@chello.nl
LKML-Reference: <20090414033936.GA534@in.ibm.com>
Signed-off-by: Ingo Molnar <mingo@elte.hu>
This commit is contained in:
Gautham R Shenoy 2009-04-14 09:09:36 +05:30 committed by Ingo Molnar
parent 132380a06b
commit 381512cf3d

View File

@ -7367,8 +7367,12 @@ static int sched_domain_debug_one(struct sched_domain *sd, int cpu, int level,
cpumask_or(groupmask, groupmask, sched_group_cpus(group));
cpulist_scnprintf(str, sizeof(str), sched_group_cpus(group));
printk(KERN_CONT " %s (__cpu_power = %d)", str,
group->__cpu_power);
printk(KERN_CONT " %s", str);
if (group->__cpu_power != SCHED_LOAD_SCALE) {
printk(KERN_CONT " (__cpu_power = %d)",
group->__cpu_power);
}
group = group->next;
} while (group != sd->groups);