On Mon, Feb 24, 2020 at 01:29:09AM +0000, Christoph Lameter wrote:
> On Sat, 22 Feb 2020, Wen Yang wrote:
Hello, Christopher!
>
> > We also observed that in this scenario, CONFIG_SLUB_CPU_PARTIAL is turned
> > on by default, and count_partial() is useless because the returned number
> > is far from the reality.
>
> Well its not useless. Its just not counting the partial objects in per cpu
> partial slabs. Those are counted by a different counter it.
Do you mean CPU_PARTIAL_ALLOC or something else?
"useless" isn't the most accurate wording, sorry for that.
The point is that the number of active objects displayed in /proc/slabinfo
is misleading if percpu partial lists are used. So it's strange to pay
for it by potentially slowing down concurrent allocations.
>
> > Therefore, we can simply return 0, then nr_free is also 0, and eventually
> > active_objects == total_objects. We do not introduce any regression, and
> > it's preferable to show the unrealistic uniform 100% slab utilization
> > rather than some very high but incorrect value.
>
> I suggest that you simply use the number of partial slabs and multiply
> them by the number of objects in a slab and use that as a value. Both
> values are readily available via /sys/kernel/slab/<...>/
So maybe something like this?
@@ -5907,7 +5907,9 @@ void get_slabinfo(struct kmem_cache *s, struct slabinfo *sinfo)
for_each_kmem_cache_node(s, node, n) {
nr_slabs += node_nr_slabs(n);
nr_objs += node_nr_objs(n);
+#ifndef CONFIG_SLUB_CPU_PARTIAL
nr_free += count_partial(n, count_free);
+#endif
}
sinfo->active_objs = nr_objs - nr_free;
Thank you!
