Messages in this thread | | | Date | Thu, 21 Mar 2013 15:34:42 +0000 | From | Mel Gorman <> | Subject | Re: [PATCH 02/10] mm: vmscan: Obey proportional scanning requirements for kswapd |
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On Thu, Mar 21, 2013 at 04:07:55PM +0100, Michal Hocko wrote: > > > > diff --git a/mm/vmscan.c b/mm/vmscan.c > > > > index 4835a7a..182ff15 100644 > > > > --- a/mm/vmscan.c > > > > +++ b/mm/vmscan.c > > > > @@ -1815,6 +1815,45 @@ out: > > > > } > > > > } > > > > > > > > +static void recalculate_scan_count(unsigned long nr_reclaimed, > > > > + unsigned long nr_to_reclaim, > > > > + unsigned long nr[NR_LRU_LISTS]) > > > > +{ > > > > + enum lru_list l; > > > > + > > > > + /* > > > > + * For direct reclaim, reclaim the number of pages requested. Less > > > > + * care is taken to ensure that scanning for each LRU is properly > > > > + * proportional. This is unfortunate and is improper aging but > > > > + * minimises the amount of time a process is stalled. > > > > + */ > > > > + if (!current_is_kswapd()) { > > > > + if (nr_reclaimed >= nr_to_reclaim) { > > > > + for_each_evictable_lru(l) > > > > + nr[l] = 0; > > > > + } > > > > + return; > > > > > > Heh, this is nicely cryptically said what could be done in shrink_lruvec > > > as > > > if (!current_is_kswapd()) { > > > if (nr_reclaimed >= nr_to_reclaim) > > > break; > > > } > > > > > > > Pretty much. At one point during development, this function was more > > complex and it evolved into this without me rechecking if splitting it > > out still made sense. > > > > > Besides that this is not memcg aware which I think it would break > > > targeted reclaim which is kind of direct reclaim but it still would be > > > good to stay proportional because it starts with DEF_PRIORITY. > > > > > > > This does break memcg because it's a special sort of direct reclaim. > > > > > I would suggest moving this back to shrink_lruvec and update the test as > > > follows: > > > > I also noticed that we check whether the scan counts need to be > > normalised more than once > > I didn't mind this because it "disqualified" at least one LRU every > round which sounds reasonable to me because all LRUs would be scanned > proportionally.
Once the scan count for one LRU is 0 then min will always be 0 and no further adjustment is made. It's just redundant to check again.
> E.g. if swappiness is 0 then nr[anon] would be 0 and > then the active/inactive aging would break? Or am I missing something? >
If swappiness is 0 and nr[anon] is zero then the number of pages to scan from every other LRU will never be adjusted. I do not see how this would affect active/inactive scanning but maybe I'm misunderstanding you.
-- Mel Gorman SUSE Labs
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