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Message-ID: <20080411162752.GB29878@linux.vnet.ibm.com>
Date:	Fri, 11 Apr 2008 09:27:52 -0700
From:	"Paul E. McKenney" <paulmck@...ux.vnet.ibm.com>
To:	Pavel Emelyanov <xemul@...nvz.org>
Cc:	Linux Netdev List <netdev@...r.kernel.org>,
	Daniel Lezcano <dlezcano@...ibm.com>,
	Denis Lunev <den@...nvz.org>,
	Linux Containers <containers@...ts.osdl.org>,
	Benjamin Thery <benjamin.thery@...l.net>
Subject: Re: [PATCH 3/14][TUN]: Introduce the tun_net structure.

On Fri, Apr 11, 2008 at 07:45:06PM +0400, Pavel Emelyanov wrote:
> Paul E. McKenney wrote:
> > On Fri, Apr 11, 2008 at 11:55:59AM +0400, Pavel Emelyanov wrote:
> >> Paul E. McKenney wrote:
> >>> On Thu, Apr 10, 2008 at 07:06:24PM +0400, Pavel Emelyanov wrote:
> >>>> This is the first step in making tuntap devices work in net 
> >>>> namespaces. The structure mentioned is pointed by generic
> >>>> net pointer with tun_net_id id, and tun driver fills one on 
> >>>> its load. It will contain only the tun devices list.
> >>>>
> >>>> So declare this structure and introduce net init and exit hooks.
> >>> OK, I have to ask...  What prevents someone else from invoking
> >>> net_generic() concurrently with a call to tun_exit_net(), potentially
> >>> obtaining a pointer to the structure that tun_exit_net() is about
> >>> to kfree()?
> >> It's the same as if the tun_net was directly pointed by the struct 
> >> net. Nobody can grant, that the pointer got by you from the struct
> >> net is not going to become free, unless you provide this security
> >> by yourself.
> > 
> > So tun_net acquires some lock before calling net_generic(), and that
> > same lock is held when calling tun_exit_net()?  Or is there but a
> 
> No.
> 
> > single tun_net task, so that it will never call tun_net_exit()
> > at the same time that it calls net_generic() for the tun_net pointer?
> 
> tun_net_exit is called only when a struct net is no longer referenced
> and is going to be kfree-ed itself, so it's impossible (or BUGy by its
> own) that someone still has a pointer on this net.
> 
> Providing the struct net is alive (!), the net->gen array is alive (or
> is scheduled for kfree after RCU grace period). Thus, if your code 
> holds the net and uses the net_generic() call, then it will get alive 
> net->gen array and alive tun_net pointer.
> 
> Next, what happens after net_generic() completes and leaves the RCU-read 
> section? Simple - the struct net is (should be) still referenced, so the
> tun_net_exit cannot yet be called and thus the tun_net pointer obtained
> earlier is alive. Unlike the (possibly) former instance of the net_generic
> array, but nobody references this one in my code (and should not do so,
> hm... I think I'll add this rule to the comments).
> 
> >> But if you call net_generic to get some pointer other than tun_net,
> >> then you're fine (due to RCU), providing you play the same rules with
> >> the pointer you're getting.
> > 
> > Agreed, RCU protects the net_generic structure, but not the structures
> > pointed to by that structure.
> 
> They are protected by struct net reference counting.

Ah, OK, got it!  Thank you for the tutorial!

> >> Maybe I'm missing something in your question, can you provide some
> >> testcase, that you suspect may cause an OOPS?
> > 
> > Just trying to understand what prevents one task from calling
> > net_generic() to pick up the tun_net pointer at the same time some other
> > task calls tun_net_exit().
> 
> If this task dereferences a "held" struct net, then should be OK. If 
> this task does not, this will OOPs in any case.
> 
> Consider the struct net to look like
> 
> struct net {
> 	...
> 	void *ptrs[N];
> }
> 
> and the net_generic to be just
> 
> static inline void net_generic(struct net *net, int id)
> {
> 	BUG_ON(id >= N);
> 	return net->ptrs[id - 1];
> }
> 
> That's the same to what I propose, except for the ptrs array is on the
> RCU protected memory.

So RCU is protecting -only- the net_generic structure that net_generic()
is traversing, and the structure returned by net_generic() is protected
by a reference counter in the upper-level struct net.

If this is the approach, I am happy.  ;-)

							Thanx, Paul
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