Re: [PATCH 10/10] locking/qspinlock: Elide back-to-back RELEASE operations with smp_wmb()
From: Andrea Parri
Date: Fri Apr 06 2018 - 09:05:27 EST
On Fri, Apr 06, 2018 at 12:34:36PM +0100, Will Deacon wrote:
> On Thu, Apr 05, 2018 at 07:28:08PM +0200, Peter Zijlstra wrote:
> > On Thu, Apr 05, 2018 at 05:59:07PM +0100, Will Deacon wrote:
> > > @@ -340,12 +341,17 @@ void queued_spin_lock_slowpath(struct qspinlock *lock, u32 val)
> > > goto release;
> > >
> > > /*
> > > + * Ensure that the initialisation of @node is complete before we
> > > + * publish the updated tail and potentially link @node into the
> > > + * waitqueue.
> > > + */
> > > + smp_wmb();
> > Maybe an explicit note to where the matching barrier lives..
> Oh man, that's not a simple thing to write: there isn't a matching barrier!
> Instead, we rely on dependency ordering for two cases:
> * We access a node by decoding the tail we get back from the xchg
> - or -
> * We access a node by following our own ->next pointer
> I could say something like:
> "Pairs with dependency ordering from both xchg_tail and explicit
> dereferences of node->next"
> but it's a bit cryptic :(
Agreed. ;) It might be helpful to instead include a snippet to highlight
the interested memory accesses/dependencies; IIUC,
* Pairs with dependency ordering from both xchg_tail and explicit/?
* dereferences of node->next:
* /* get node0, encode node0 in tail */
* ((struct pv_node *)node0)->cpu = smp_processor_id();
* ((struct pv_node *)node0)->state = vcpu_running;
* old = xchg_tail(lock, tail);
* /* get node1, encode tail from node1 */
* old = xchg_tail(lock, tail); // = tail corresponding to node0
* // head an addr. dependency
* /* decode old in prev */
* pv_wait_node(node1, prev);
* READ ((struct pv_node *)prev)->cpu // addr. dependent read
* READ ((struct pv_node *)prev)->state // addr. dependend read
* [More details for the case "following our own ->next pointer" you
* mentioned dabove.]
CPU1 would also have:
WRITE_ONCE(prev->next, node1); // addr. dependent write
but I'm not sure how this pairs: does this belong to the the second
case above? can you elaborate on that?