The interval_tree_subtree_search() holds the loop invariant:
start <= node->ITSUBTREE
Let's say we have a following tree:
node
/ \
left right
So we know node->ITSUBTREE is contributed by one of the following:
* left->ITSUBTREE
* ITLAST(node)
* right->ITSUBTREE
When we come to the right node, we are sure the first two don't
contribute to node->ITSUBTREE and it must be the right node does the
job.
So skip the check before go to the right subtree.
Signed-off-by: Wei Yang <richard.weiyang@xxxxxxxxx>
CC: Matthew Wilcox <willy@xxxxxxxxxxxxx>
CC: Michel Lespinasse <michel@xxxxxxxxxxxxxx>
---
include/linux/interval_tree_generic.h | 8 ++------
1 file changed, 2 insertions(+), 6 deletions(-)
diff --git a/include/linux/interval_tree_generic.h b/include/linux/interval_tree_generic.h
index aaa8a0767aa3..1b400f26f63d 100644
--- a/include/linux/interval_tree_generic.h
+++ b/include/linux/interval_tree_generic.h
@@ -104,12 +104,8 @@ ITPREFIX ## _subtree_search(ITSTRUCT *node, ITTYPE start, ITTYPE last) \
if (ITSTART(node) <= last) { /* Cond1 */ \
if (start <= ITLAST(node)) /* Cond2 */ \
return node; /* node is leftmost match */ \
- if (node->ITRB.rb_right) { \
- node = rb_entry(node->ITRB.rb_right, \
- ITSTRUCT, ITRB); \
- if (start <= node->ITSUBTREE) \
- continue; \
- } \
+ node = rb_entry(node->ITRB.rb_right, ITSTRUCT, ITRB); \
+ continue; \
} \
return NULL; /* No match */ \
} \
--
2.34.1
