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Message-ID: <20070306094149.GC12365@wotan.suse.de>
Date:	Tue, 6 Mar 2007 10:41:49 +0100
From:	Nick Piggin <npiggin@...e.de>
To:	Eric Dumazet <dada1@...mosbay.com>
Cc:	David Miller <davem@...emloft.net>, netdev@...r.kernel.org,
	robert.olsson@....uu.se, Paul McKenney <paulmck@...ibm.com>
Subject: Re: [RFC PATCH]: Dynamically sized routing cache hash table.

On Tue, Mar 06, 2007 at 10:23:44AM +0100, Eric Dumazet wrote:
> On Tuesday 06 March 2007 10:11, Nick Piggin wrote:
> 
> > Cool! I have some fixes for the rcu barrier issues, with some C-style
> > comments and questions :)
> >
> > I was going to send you a fix first for the rcu barriers, then a
> > second to convert the read-side to a barrier-less one that I described,
> > however considering that your patch is a WIP in progress anyway, I
> > won't worry too much about the normal protocol.
> >
> > I _think_ my reasoning regarding the rcu barriers and grace periods
> > is correct. I'll keep thinking about it though. (Paul cc'ed).
> >
> > I'm not so familiar with this code, so I have sprinkled around a lot
> > of comments that could be pure crap ;) They are mainly just to help
> > you ensure that you cover all bases... compile tested only at this
> > stage.
> 
> I think we missed :
> 
> +static void rt_hash_resize_work(struct work_struct *work)
> 
> +
> +			*head = rth->u.dst.rt_next;
> +
> +			hash = rt_hashfn(rt_hash,
> +					 rth->fl.fl4_dst,
> +					 rth->fl.fl4_src,
> +					 iface);
> +			rth->u.dst.rt_next = rt_hash->table[hash].chain;
> +			rt_hash->table[hash].chain = rth;
> 
> This really needs some ..._del_rcu()/..._add_rcu()_ ... primitives, no ?
> Or else a reader might be very confused...

I'm not sure... this code really depends on the hash table management,
rather than the management of the hash tables, if you understand me ;)

>From what I can _see_, this is similar to how rt_intern_hash does it.
I don't know exactly why rt_intern_hash can get away without using
rcu_assign_pointer in some cases, however:

Note that we don't need an rcu_assign_pointer for this, because the
memory operations that initialized the entry have already been ordered
when it was first inserted.

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