Hi Peter~
Sorry about the confusing log, and thanks for the patient. :)
Here, I want to say something more about the sched_domains_numa_levels
to make myself more clear. :)
Let's have an example here.
sched_init_numa()
{
...
// A loop set sched_domains_numa_levels to level.-------------1
// I set sched_domains_numa_levels to 0.
sched_domains_numa_levels = 0;--------------------------------2
// A loop allocating memory for sched_domains_numa_masks[][]
for (i = 0; i < level; i++) {
......
// Allocate memory for sched_domains_numa_masks[i]----3
......
}
......
// I reset sched_domains_numa_levels to level.
sched_domains_numa_levels = level;----------------------------4
}
// A new function I added.
static void sched_domains_numa_masks_clear(int cpu)
{
int i, j;
for (i = 0; i < sched_domains_numa_levels; i++)---------------5
for (j = 0; j < nr_node_ids; j++)
cpumask_clear_cpu(cpu,
sched_domains_numa_masks[i][j]);
}
Suppose level is 10, and at step 1, sched_domains_numa_levels is 10.
If I didn't set sched_domains_numa_levels to 0 at step 2, it will be 10
all the time.
If memory allocation at step 3 fails when i = 5, the array
sched_domains_numa_masks[][] will only have 5 members, and
sched_domains_numa_levels is 10.
As you see, I added 2 functions using sched_domains_numa_levels to
iterate sched_domains_numa_masks[][], such as at step 5.
In this case, the iteration will break out when i = 5.
This could be dangerous.
So, I set sched_domains_numa_levels to 0 at step 2. This way, even if
any memory allocation at step 3 fails, and sched_init_numa() returns,
anyone uses sched_domains_numa_levels (which is 0) won't be wrong.
I'm not sure if this is the best way to settle this problem.
If you have a better idea, please tell me. Thanks. :)
On 09/25/2012 06:33 PM, Peter Zijlstra wrote:
On Tue, 2012-09-25 at 10:39 +0800, Tang Chen wrote:
@@ -6765,11 +6773,64 @@ static void sched_init_numa(void)
}
sched_domain_topology = tl;
+
+ sched_domains_numa_levels = level;
And I set it to level here again.
But its already set there.. its set every time we find a new level.
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