Hi Martin,
On going thru the code, I could think of this answer to your
question.(unfortunately my kernel version is 2.4.2 but I guess
logic may still hold good; anyway) The function try_to_swap_out
gets called ultimately from refill_inactive() which goes thru the
list of inactive pages and flushes them to swap. refill_inactive()
is called from do_try_to_free_pages() but before that page_launder
is invoked. page launder makes sure that the pages in inactive
clean list, swap cache and disk are in sync. So after that when
try_to_swap_out() gets called, it need not bother about dirty
pages.
But I have my own doubts on swapping which I would like to get
cleared. I am unable to get the reason why at all a shared page
gets transferred to the disk as long as it is in use.
Why won't the following steps work:
(i) In case the page is shared and the 1st time try_to_swap_out()
is called : the page is transferred to swap cache and
__free_page() is called the page count is not zero. Then do not
transfer the page to disk.
(ii) When the last process that shared the page calls
try_to_swap_out: the pagecount hits 0. Then transfer the page to
disk.
This way for shared pages only one disk transer(which is
expensive) gets done for shared pages.
My 2 bits
Raghav
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