netdev - Re: [PATCH net-next] sch_netem: faster rb tree removal

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Date:   Mon, 25 Sep 2017 10:14:23 -0600
From:   David Ahern <dsahern@...il.com>
To:     Eric Dumazet <eric.dumazet@...il.com>
Cc:     David Miller <davem@...emloft.net>,
        netdev <netdev@...r.kernel.org>,
        Stephen Hemminger <stephen@...workplumber.org>
Subject: Re: [PATCH net-next] sch_netem: faster rb tree removal

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 <dsahern@...il.com>