On Sun, 2017-09-24 at 20:05 -0600, David Ahern wrote:
> On 9/24/17 7:57 PM, David Ahern wrote:
> > Hi Eric:
> >
> > I'm guessing the cost is in the rb_first and rb_next computations. Did
> > you consider something like this:
> >
> > struct rb_root *root
> > struct rb_node **p = &root->rb_node;
> >
> > while (*p != NULL) {
> > struct foobar *fb;
> >
> > fb = container_of(*p, struct foobar, rb_node);
> > // fb processing
> rb_erase(&nh->rb_node, root);
>
> > p = &root->rb_node;
> > }
> >
>
> Oops, dropped the rb_erase in my consolidating the code to this snippet.
Hi David
This gives about same numbers than method_1
I tried with 10^7 skbs in the tree :
Your suggestion takes 66ns per skb, while the one I chose takes 37ns per
skb.
Thanks.
