On Sun, Jun 09, 2019 at 05:10:28PM +0800, ChenGang wrote: >Usually the value of min_free_kbytes is multiply of 4, >and in this case ,the right shift is ok. >But if it's not, the right-shifting operation will lose the low 2 bits,
But PAGE_SHIFT is not always 12. >and this cause kernel don't reserve enough memory. >So it's necessary to align the value of min_free_kbytes to multiply of 4. >For example, if min_free_kbytes is 64, then should keep 16 pages, >but if min_free_kbytes is 65 or 66, then should keep 17 pages. > >Signed-off-by: ChenGang <[email protected]> >--- > mm/page_alloc.c | 3 ++- > 1 file changed, 2 insertions(+), 1 deletion(-) > >diff --git a/mm/page_alloc.c b/mm/page_alloc.c >index d66bc8a..1baeeba 100644 >--- a/mm/page_alloc.c >+++ b/mm/page_alloc.c >@@ -7611,7 +7611,8 @@ static void setup_per_zone_lowmem_reserve(void) > > static void __setup_per_zone_wmarks(void) > { >- unsigned long pages_min = min_free_kbytes >> (PAGE_SHIFT - 10); >+ unsigned long pages_min = >+ (PAGE_ALIGN(min_free_kbytes * 1024) / 1024) >> (PAGE_SHIFT - >10); In my mind, pages_min is an estimated value. Do we need to be so precise? > unsigned long lowmem_pages = 0; > struct zone *zone; > unsigned long flags; >-- >1.8.5.6 -- Wei Yang Help you, Help me
