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,
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 <cg.chen@xxxxxxxxxx>
---
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);
unsigned long lowmem_pages = 0;
struct zone *zone;
unsigned long flags;
--
1.8.5.6
