At 06/01/2016 12:23 AM, David Sterba wrote:
On Tue, May 31, 2016 at 03:27:42PM +0800, luke wrote:
+static void ref_root_fini(struct ref_root *ref_tree)
+{
+ struct ref_node *node;
+ struct rb_node *next;
+
+ while ((next = rb_first(&ref_tree->rb_root)) != NULL) {
+ node = rb_entry(next, struct ref_node, rb_node);
+ rb_erase(next, &ref_tree->rb_root);
+ kfree(node);
This could be slow as rb_erase has to do the rb-tree rotations. Can we
do a post-order traversal and just free the nodes?
Excuse me, this seems to be standard way to release the rb-tree. For
example, "btrfs_free_block_groups" this function is the same way. So
should i refactor this function?
So for first version it's ok to keep it the simple way, but refactoring
to the post-order traversal should help performance. And as there are
more places it would be better to do it separately.
Yes, you're right. I have updated this patch to fix other problems. And,
I'll refactor this function with another patch soon.
Thanks,
Lu
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