Re: memory barrier question
From: James Bottomley
Date: Thu Sep 16 2010 - 13:40:51 EST
On Thu, 2010-09-16 at 19:17 +0200, Miklos Szeredi wrote:
> On Thu, 16 Sep 2010, James Bottomley wrote:
> > On Thu, 2010-09-16 at 18:56 +0200, Miklos Szeredi wrote:
> > > On Thu, 16 Sep 2010, Paul E. McKenney wrote:
> > > > On Thu, Sep 16, 2010 at 06:06:53PM +0200, Miklos Szeredi wrote:
> > > > > On Thu, 16 Sep 2010, Paul E. McKenney wrote:
> > > > > > On Thu, Sep 16, 2010 at 03:30:56PM +0100, David Howells wrote:
> > > > > > > Miklos Szeredi <miklos@xxxxxxxxxx> wrote:
> > > > > > >
> > > > > > > > Is the rmb() really needed?
> > > > > > > >
> > > > > > > > Take this code from fs/namei.c for example:
> > > > > > > >
> > > > > > > > inode = next.dentry->d_inode;
> > > > > > > > if (!inode)
> > > > > > > > goto out_dput;
> > > > > > > >
> > > > > > > > if (inode->i_op->follow_link) {
> > > > > > > >
> > > > > > > > It happily dereferences dentry->d_inode without a barrier after
> > > > > > > > checking it for non-null, while that d_inode might have just been
> > > > > > > > initialized on another CPU with a freshly created inode. There's
> > > > > > > > absolutely no synchornization with that on this side.
> > > > > > >
> > > > > > > Perhaps it's not necessary; once set, how likely is i_op to be changed once
> > > > > > > I_NEW is cleared?
> > > > > >
> > > > > > Are the path_get()s protecting this?
> > > > >
> > > > > No, when creating a file the dentry will go from negative to positive
> > > > > independently from lookup. The dentry can get instantiated with an
> > > > > inode between the path_get() and dereferencing ->d_inode.
> > > > >
> > > > > > If there is no protection, then something like rcu_dereference() is
> > > > > > needed for the assignment from next.dentry->d_inode.
> > > > >
> > > > > Do I understand correctly that the problem is that a CPU may have a
> > > > > stale cache associated with *inode, one that was loaded before the
> > > > > write barrier took effect?
> > > >
> > > > Yes, especially if the compiler is aggressively optimizing.
> > >
> > > How do compiler optimizations make a difference?
> >
> > There are two types of reorderings that cause problems if you expect the
> > bus visible ordering to matter. One is CPU issue reordering, where the
> > cpu decides to output loads and stores in a different order than the
> > input instruction stream actually said. The other is compiler
> > re-ordering where the compiler actually reorders the instructions to
> > execute in a different order from what you'd expect by simply reading
> > the C code.
> >
> > We have compiler barrier instructions for the latter and barriers which
> > issue CPU primitives for the former.
>
> Right but in the concrete namei example I can't see how a compiler
> optimization can make a difference. The order of the loads is quite
> clear:
>
> LOAD inode = next.dentry->inode
> if (inode != NULL)
> LOAD inode->f_op
>
> What is there the compiler can optimize?
In this example, it can't the if is conditional on the previous
executions, so even a speculating CPU is required to do ordering If the
compiler could prove inode wasn't null (which it can't I think in this
case) it might then re-order. in your original question, there were
ways the compiler could behave strangely (for instance, it would likely
start by initialising p to &x rather than NULL).
James
--
To unsubscribe from this list: send the line "unsubscribe linux-kernel" in
the body of a message to majordomo@xxxxxxxxxxxxxxx
More majordomo info at http://vger.kernel.org/majordomo-info.html
Please read the FAQ at http://www.tux.org/lkml/