sched: Avoid printing sched_group::__cpu_power for default case
Gautham R Shenoy [Tue, 14 Apr 2009 03:39:36 +0000 (08:39 +0530)]
Commit 46e0bb9c12f4 ("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>

kernel/sched.c

index e90e70e..b902e58 100644 (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);