On Thu, Sep 29, 2016 at 03:36:38PM -0400, Alan Stern wrote: > > > If you execute P0 and P1 concurrently and one store of each 'wins' the > > > LWSYNC of either is null and void, and therefore P2 is unordered and can > > > observe r2==0. > > Not so. lwsync instructions cannot be "voided".
I distinctly remember there being a case (smp_wmb()) where the lwsync would disappear if the store was lost.
