Subject: Re: [PATCH] net: No more expensive sock_hold()/sock_put() on each tx

From: Eric Dumazet <[email-masked]>
Date: Wed, 10 Jun 2009 10:30:49 +0200

David Miller a écrit :
> From: Eric Dumazet <[email-masked]>
> Date: Thu, 04 Jun 2009 11:18:35 +0200
> 
>> @@ -1172,12 +1186,18 @@ void __init sk_init(void)
>>  void sock_wfree(struct sk_buff *skb)
>>  {
>>  	struct sock *sk = skb->sk;
>> +	int res;
>>  
>>  	/* In case it might be waiting for more memory. */
>> -	atomic_sub(skb->truesize, &sk->sk_wmem_alloc);
>> +	res = atomic_sub_return(skb->truesize, &sk->sk_wmem_alloc);
>>  	if (!sock_flag(sk, SOCK_USE_WRITE_QUEUE))
>>  		sk->sk_write_space(sk);
>> -	sock_put(sk);
>> +	/*
>> +	 * if sk_wmem_alloc reached 0, we are last user and should
>> +	 * free this sock, as sk_free() call could not do it.
>> +	 */
>> +	if (res == 0)
>> +		__sk_free(sk);
>>  }
>>  EXPORT_SYMBOL(sock_wfree);
>>  
> 
> Eric, I don't understand this part, please enlighten me :-)
> 
> Just because we've liberated all of the write buffer space, that does
> not mean that it's time to kill off the socket completely.
> 
> Right?

Remember we initialize this field to one.

If we freed all write buffer space, final value is one, not zero.

res == 0 only if we both freed all write buffer space, *and* socket was
also refcounted to 0 (sk_free() then realized it could not yet call __sk_free())

So we cheat a litle bit, because of this offset of one, we might block a sender a litle bit earlier :)

Thank you