netdev - Re: [PATCH net-next] net: move inet_dport/inet_num in sock

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Date:	Wed, 28 Nov 2012 03:12:31 +0000
From:	Ben Hutchings <bhutchings@...arflare.com>
To:	Joe Perches <joe@...ches.com>
CC:	Eric Dumazet <eric.dumazet@...il.com>,
	David Miller <davem@...emloft.net>,
	netdev <netdev@...r.kernel.org>,
	Ling Ma <ling.ma.program@...il.com>
Subject: Re: [PATCH net-next] net: move inet_dport/inet_num in sock_common

On Tue, 2012-11-27 at 18:23 -0800, Joe Perches wrote:
> On Tue, 2012-11-27 at 13:24 -0800, Eric Dumazet wrote:
> > On Tue, 2012-11-27 at 09:23 -0800, Joe Perches wrote:
> > > On Tue, 2012-11-27 at 07:06 -0800, Eric Dumazet wrote:
> > > > From: Eric Dumazet <edumazet@...gle.com>
> > > > 
> > > > commit 68835aba4d9b (net: optimize INET input path further)
> > > > moved some fields used for tcp/udp sockets lookup in the first cache
> > > > line of struct sock_common.
> > > []
> > > > diff --git a/include/linux/ipv6.h b/include/linux/ipv6.h
> > > > index 5e11905..196ede4 100644
> > > > --- a/include/linux/ipv6.h
> > > > +++ b/include/linux/ipv6.h
> > > > @@ -365,19 +365,21 @@ static inline struct raw6_sock *raw6_sk(const struct sock *sk)
> > > >  #endif /* IS_ENABLED(CONFIG_IPV6) */
> > > >  
> > > >  #define INET6_MATCH(__sk, __net, __hash, __saddr, __daddr, __ports, __dif)\
> > > > +	(((__sk)->sk_hash == (__hash)) &&					\
> > > > +	 ((*((__portpair *)&(inet_sk(__sk)->inet_dport))) == (__ports)) &&	\
> > > > +	 ((__sk)->sk_family		== AF_INET6)		&&		\
> > > 
> > > Perhaps these could be |'d together to avoid the test/jump
> > > after each comparison by using some bit operations instead.
> > > 
> > > > +	 ipv6_addr_equal(&inet6_sk(__sk)->daddr, (__saddr))	&&		\
> > > > +	 ipv6_addr_equal(&inet6_sk(__sk)->rcv_saddr, (__daddr))	&&		\
> > > > +	 (!((__sk)->sk_bound_dev_if) || ((__sk)->sk_bound_dev_if == (__dif))) && \
> > > > +	 net_eq(sock_net(__sk), (__net)))
> > > 
> > But it would be wrong.
> 
> OK, so it's an and not an or.  Duh.
[...]

The way to combine these sorts of comparisons is along the lines of:

(((left->a ^ right->a) |
  (left->b ^ right->b) |
  ...) == 0)

But when there are big-endian types involved, sparse is likely to
complain about combining them.

Ben.

-- 
Ben Hutchings, Staff Engineer, Solarflare
Not speaking for my employer; that's the marketing department's job.
They asked us to note that Solarflare product names are trademarked.

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