On 9/24/17 11:27 PM, Eric Dumazet wrote:
> 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 for the test.
I made a simple program this morning and ran it under perf. With the
above suggestion the rb_erase has a high cost because it always deletes
the root node. Your method 1 has a high cost on rb_first which is
expected given its definition and it is run on each removal. Both
options increase in time with the number of entries in the tree.
Your method 2 is fairly constant from 10,000 entries to 10M entries
which makes sense: a one time cost at finding rb_first and then always
removing a bottom node so rb_erase is light.
As for the change:
Acked-by: David Ahern <dsah...@gmail.com>
