On 10/14/07, shyam Burkule <shyam.burkule@xxxxxxxxx> wrote:
On 10/13/07, Peter Teoh <htmldeveloper@xxxxxxxxx > wrote:I think I have understood better.
On 10/13/07, shyam Burkule < shyam.burkule@xxxxxxxxx> wrote:
> Hi
> Now as given in fig. if there is second reference mark_page_accessed()
> will invoke and then change flag PG_referenced to 1 and still page is in
> inacive list.Is it really true...i mean at this point of reference there
> will be cache hit and therefore controll doesnt comes to kernel.
> And also as i think this set of PG_referenced is very important.So if it is
> not true then what and how LRU moves pages across its inactive and active
> list.
See:
http://kernel.org/doc/gorman/html/understand/understand013.html
Ok, from diagram, inactive->active is done in shrink_cache().This is point of question ....When page is in inactive list and if any process trying to access page ,there will be cache hit.And In practice we simple cannot modify LRU list on cache hit...so is it really true that then function() is get called when there is reference to page,when it is in inactive list?? If not then when page is moved from inactive to active list?.
active->inactive is done in refill_inactive().
Please read in depth 10.3.2 and 10.3.1. ALl the answer is there.
In particular this line:
>>>From there, the function is basically a very large for-loop which
scans at most max_scan pages to free up nr_pages pages from the end of
the inactive_list or until the inactive_list is empty. After each
page, it checks to see whether it should reschedule itself so that the
swapper does not monopolise the CPU.
And so I think the timing in non-deterministic, because if schedule()
can be called after every page move from inactive_list to active_list,
then there is no timing at all.
--
Shyam
"Born To Learn"
--
Shyam
"Born To Learn"
