On Fri, Jun 17, 2016 at 11:26:41AM -0400, Waiman Long wrote:
> On 06/16/2016 08:48 PM, Boqun Feng wrote:
> >On Thu, Jun 16, 2016 at 05:35:54PM -0400, Waiman Long wrote:
> >>If you look into the actual code:
> >>
> >> next = xchg_release(&node->next, NULL);
> >> if (next) {
> >> WRITE_ONCE(next->locked, 1);
> >> return;
> >> }
> >>
> >>There is a control dependency that WRITE_ONCE() won't happen until
> >But a control dependency only orders LOAD->STORE pairs, right? And here
> >the control dependency orders the LOAD part of xchg_release() and the
> >WRITE_ONCE().
> >
> >Along with the fact that RELEASE only orders the STORE part of xchg with
> >the memory operations preceding the STORE part, so for the following
> >code:
> >
> > WRTIE_ONCE(x,1);
> > next = xchg_release(&node->next, NULL);
> > if (next)
> > WRITE_ONCE(next->locked, 1);
> >
> >such a reordering is allowed to happen on ARM64v8
> >
> > next = ldxr [&node->next] // LOAD part of xchg_release()
> >
> > if (next)
> > WRITE_ONCE(next->locked, 1);
> >
> > WRITE_ONCE(x,1);
> > stlxr NULL [&node->next] // STORE part of xchg_releae()
> >
> >Am I missing your point here?
>
> My understanding of the release barrier is that both prior LOADs and STOREs
> can't move after the barrier. If WRITE_ONCE(x, 1) can move to below as shown
> above, it is not a real release barrier and we may need to change the
> barrier code.
You seem to be missing the point.
{READ,WRITE}_ONCE accesses appearing in program order after a release
are not externally ordered with respect to the release unless they
access the same location.
This is illustrated by Boqun's example, which shows two WRITE_ONCE
accesses being reordered before a store-release forming the write
component of an xchg_release. In both cases, WRITE_ONCE(x, 1) remains
ordered before the store-release.
Will
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