Re: [PATCH locking/Documentation 1/2] Add note of release-acquire store vulnerability

[Paul E. McKenney]

On Thu, Sep 29, 2016 at 06:17:38PM +0200, Peter Zijlstra wrote:
> On Thu, Sep 29, 2016 at 05:03:08PM +0100, Will Deacon wrote:
> > On Thu, Sep 29, 2016 at 05:58:17PM +0200, Peter Zijlstra wrote:
> > > On Thu, Sep 29, 2016 at 08:54:01AM -0700, Paul E. McKenney wrote:
> > > > If two processes are related by a RELEASE+ACQUIRE pair, ordering can be
> > > > broken if a third process overwrites the value written by the RELEASE
> > > > operation before the ACQUIRE operation has a chance of reading it.
> > > > This commit therefore updates the documentation to call this vulnerability
> > > > out explicitly.
> > > > 
> > > > Reported-by: Alan Stern <stern@xxxxxxxxxxxxxxxxxxx>
> > > > Signed-off-by: Paul E. McKenney <paulmck@xxxxxxxxxxxxxxxxxx>
> > > 
> > > > +     However, please note that a chain of RELEASE+ACQUIRE pairs may be
> > > > +     broken by a store by another thread that overwrites the RELEASE
> > > > +     operation's store before the ACQUIRE operation's read.
> > > 
> > > This is the powerpc lwsync quirk, right? Where the barrier disappears
> > > when it looses the store.
> > > 
> > > Or is there more to it? Its not entirely clear from the Changelog, which
> > > I feel should describe the reason for the behaviour.
> > 
> > If I've groked it correctly, it's for cases like:
> > 
> > 
> > PO:
> > Wx=1
> > WyRel=1
> > 
> > P1:
> > Wy=2
> > 
> > P2:
> > RyAcq=2
> > Rx=0
> > 
> > Final value of y is 2.
> > 
> > 
> > This is permitted on arm64. If you make P1's store a store-release, then
> > it's forbidden, but I suspect that's not generally true of the kernel
> > memory model.
> 
> Right, I think that on PowerPC, even if P1 does store-release you can
> still get this, since the two stores conflict one can loose out, and the
> lwsync associated with the loosing store gets removed along with it.
> 
> So yes, I think this needs more clarification.

Whether the store is loose or not, yes, putting an lwsync as the first
instruction in a given task has no effect.  ;-)

							Thanx, Paul