Message-ID: <48EB7747.9060505@cosmosbay.com>
Date: Tue, 07 Oct 2008 16:50:47 +0200
From: Eric Dumazet <dada1@cosmosbay.com>
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To: Christoph Lameter <cl@linux-foundation.org>
Cc: Peter Zijlstra <a.p.zijlstra@chello.nl>, minyard@acm.org,
       Linux Kernel <linux-kernel@vger.kernel.org>, netdev@vger.kernel.org,
       shemminger@vyatta.com, paulmck@linux.vnet.ibm.com
Subject: Re: [PATCH 3/3] Convert the UDP hash lock to RCU
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Christoph Lameter a ?crit :
> Eric Dumazet wrote:
>>>> Or just add SLAB_DESTROY_BY_RCU to slab creation in proto_register()
>>>> for "struct proto udp_prot/udpv6_prot" so that kmem_cache_free() done
>>>> in sk_prot_free() can defer freeing to RCU...
>>> Be careful!, SLAB_DESTROY_BY_RCU just means the slab page gets
>>> RCU-freed, this means that slab object pointers stay pointing to valid
>>> memory, but it does _NOT_ mean those slab objects themselves remain
>>> valid.
>>>
>>> The slab allocator is free to re-use those objects at any time -
>>> irrespective of the rcu-grace period. Therefore you will have to be able
>>> to validate that the object you point to is indeed the object you
>>> expect, otherwise strange and wonderful things will happen.
>>>
>> Thanks for this clarification. I guess we really need a rcu head then :)
> 
> No you just need to make sure that the object you located is still active
> (f.e. refcount > 0) and that it is really a match (hash pointers may be
> updated asynchronously and therefore point to the object that has been reused
> for something else).
> 
> Generally it is advisable to use SLAB_DESTROY_BY_RCU because it preserves the
> cache hot advantages of the objects. Regular RCU freeing will let the object
> expire for a tick or so which will result in the cacheline cooling down.

Seems really good to master this SLAB_DESTROY_BY_RCU thing (I see almost no use
of it in current kernel)

1) Hum, do you know why "struct file" objects dont use SLAB_DESTROY_BY_RCU then,
since we noticed a performance regression for several workloads at RCUification
of file structures ?

2) What prevents an object to be *freed* (and deleted from a hash chain), then
re-allocated and inserted to another chain (different keys) ? (final refcount=1)

If the lookup detects a key mismatch, how will it continue to the next item,
since 'next' pointer will have been reused for the new chain insertion...

Me confused...


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