On Fri, Jul 01, 2016 at 08:08:07AM +0100, David Howells wrote:
> Should I also reorder rb_replace_node() whilst
> I'm at it so that the new node is initialised first (it shouldn't make a
> difference, I know)?
Might as well, I can't imagine that making a performance difference and
keeping the general structure of things similar helps avoid confusion.
> commit 812667d2a82a6a8fe35a44e951e8b1515b04696a
> Author: David Howells <dhowe...@redhat.com>
> Date: Fri Jul 1 07:53:51 2016 +0100
>
> Introduce rb_replace_node_rcu()
>
> Implement an RCU-safe variant of rb_replace_node().
>
> Signed-off-by: David Howells <dhowe...@redhat.com>
> cc: Peter Zijlstra <pet...@infradead.org>
One little niggle below, but:
Acked-by: Peter Zijlstra (Intel) <pet...@infradead.org>
> diff --git a/lib/rbtree.c b/lib/rbtree.c
> index 1356454e36de..59eb906c6c3b 100644
> --- a/lib/rbtree.c
> +++ b/lib/rbtree.c
> @@ -551,6 +551,25 @@ void rb_replace_node(struct rb_node *victim, struct
> rb_node *new,
> }
> EXPORT_SYMBOL(rb_replace_node);
>
> +void rb_replace_node_rcu(struct rb_node *victim, struct rb_node *new,
> + struct rb_root *root)
> +{
> + struct rb_node *parent = rb_parent(victim);
> +
> + /* Copy the pointers/colour from the victim to the replacement */
> + *new = *victim;
> +
> + /* Set the surrounding nodes to point to the replacement */
> + if (victim->rb_left)
> + rb_set_parent(victim->rb_left, new);
> + if (victim->rb_right)
> + rb_set_parent(victim->rb_right, new);
> +
> + /* Set the onward pointer last with an RCU barrier */
Maybe also explain _why_ this needs to be last. Its obvious now and to
us, but it might safe some head scratching later.
> + __rb_change_child_rcu(victim, new, parent, root);
> +}
> +EXPORT_SYMBOL(rb_replace_node_rcu);
