From: David Woodhouse Date: Fri, 21 Apr 2006 12:30:36 +0000 (+0100) Subject: [RBTREE] Remove dead code in rb_erase() X-Git-Tag: firefly_0821_release~35481^2~5 X-Git-Url: http://demsky.eecs.uci.edu/git/?a=commitdiff_plain;h=1975e59375756da4ff4e6e7d12f67485e813ace0;p=firefly-linux-kernel-4.4.55.git [RBTREE] Remove dead code in rb_erase() Observe rb_erase(), when the victim node 'old' has two children so neither of the simple cases at the beginning are taken. Observe that it effectively does an 'rb_next()' operation to find the next (by value) node in the tree. That is; we go to the victim's right-hand child and then follow left-hand pointers all the way down the tree as far as we can until we find the next node 'node'. We end up with 'node' being either the same immediate right-hand child of 'old', or one of its descendants on the far left-hand side. For a start, we _know_ that 'node' has a parent. We can drop that check. We also know that if 'node's parent is 'old', then 'node' is the right-hand child of its parent. And that if 'node's parent is _not_ 'old', then 'node' is the left-hand child of its parent. So instead of checking for 'node->rb_parent == old' in one place and also checking 'node's heritage separately when we're trying to change its link from its parent, we can shuffle things around a bit and do it like this... Signed-off-by: David Woodhouse --- diff --git a/lib/rbtree.c b/lib/rbtree.c index 14b791ac5089..63473e04f18a 100644 --- a/lib/rbtree.c +++ b/lib/rbtree.c @@ -243,18 +243,13 @@ void rb_erase(struct rb_node *node, struct rb_root *root) if (child) child->rb_parent = parent; - if (parent) - { - if (parent->rb_left == node) - parent->rb_left = child; - else - parent->rb_right = child; - } - else - root->rb_node = child; - if (node->rb_parent == old) + if (node->rb_parent == old) { + parent->rb_right = child; parent = node; + } else + parent->rb_left = child; + node->rb_parent = old->rb_parent; node->rb_color = old->rb_color; node->rb_right = old->rb_right;