On 3/27/20 3:01 PM, Wei Yang wrote:
Since we always clear node_order before getting it, we can leverage
compiler to do this instead of at run time.
Signed-off-by: Wei Yang <richard.weiyang@xxxxxxxxx>
---
mm/page_alloc.c | 3 +--
1 file changed, 1 insertion(+), 2 deletions(-)
diff --git a/mm/page_alloc.c b/mm/page_alloc.c
index dfcf2682ed40..49dd1f25c000 100644
--- a/mm/page_alloc.c
+++ b/mm/page_alloc.c
@@ -5585,7 +5585,7 @@ static void build_thisnode_zonelists(pg_data_t *pgdat)
static void build_zonelists(pg_data_t *pgdat)
{
- static int node_order[MAX_NUMNODES];
+ static int node_order[MAX_NUMNODES] = {0};
Looks wrong: now the single instance of node_order is initialized just once by
the compiler. And that means that only the first caller of this function
gets a zeroed node_order array...
int node, load, nr_nodes = 0;
nodemask_t used_mask = NODE_MASK_NONE;
int local_node, prev_node;
@@ -5595,7 +5595,6 @@ static void build_zonelists(pg_data_t *pgdat)
load = nr_online_nodes;
prev_node = local_node;
- memset(node_order, 0, sizeof(node_order));
...and all subsequent callers are left with whatever debris is remaining in
node_order. So this is not good.
The reason that memset() was used here, is that there aren't many other ways
to get node_order zeroed, given that it is a statically defined variable.
while ((node = find_next_best_node(local_node, &used_mask)) >= 0) {
/*
* We don't want to pressure a particular node.
thanks,
--
John Hubbard
NVIDIA
