From mboxrd@z Thu Jan  1 00:00:00 1970
From: Eric Dumazet <eric.dumazet@gmail.com>
Subject: Re: [PATCH] NFS: Fix RCU warnings in
 nfs_inode_return_delegation_noreclaim() [ver #2]
Date: Wed, 31 Mar 2010 20:32:46 +0200
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Cc: paulmck@linux.vnet.ibm.com, Trond.Myklebust@netapp.com,
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To: David Howells <dhowells@redhat.com>
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Le mercredi 31 mars 2010 =C3=A0 18:37 +0100, David Howells a =C3=A9crit=
 :
> Paul E. McKenney <paulmck@linux.vnet.ibm.com> wrote:
>=20
> > Protected by something that the caller did, be it holding the the c=
orrect
> > lock, operating on it during initialization before other CPUs have =
access
> > to it, operating on it during cleanup after other CPUs' access has =
been
> > revoked, or whatever.
>=20
> But the point I made very early this morning still stands:  What if s=
omeone
> simply wants to test the pointer, not actually to dereference it?
>=20
> NFS was using rcu_dereference() for this in a couple of places - whic=
h is
> overkill.  I suggested stripping this off and you countered with the
> suggestion that it should be using rcu_dereference_check().
>=20

If pointer has the rcu mark, and somehing access this pointer without
proper locking, then automatic checkers (sparse...) will trigger a
warning, this is what Paul said.

Example of such checks,=20

# define __percpu   __attribute__((noderef, address_space(3)))=20

If someone tries to manipulate a __percpu marked ptr without proper API=
,
sparse loudly complains.


> Why do I need anything at all?
>=20

If you dont own a lock, and test a pointer, what guarantee do you have
this pointer doesnt change right after you tested it ?

If *something* protects the pointer from being changed, then how can be
expressed this fact ?

If nothing protects the pointer, why test it then, as result of test is
unreliable ?

If NFS was using rcu_dereference(), it probably was for a reason, but i=
f
nobody can recall it, it was a wrong reason ?

Sorry, too many questions and no answer I guess...