Re: XDP socket rings, and LKMM litmus tests

From: Alan Stern
Date: Fri Mar 05 2021 - 11:16:01 EST


On Fri, Mar 05, 2021 at 09:12:30AM +0800, Boqun Feng wrote:
> On Thu, Mar 04, 2021 at 11:11:42AM -0500, Alan Stern wrote:

> > Forget about local variables for the time being and just consider
> >
> > dep ; [Plain] ; rfi
> >
> > For example:
> >
> > A: r1 = READ_ONCE(x);
> > y = r1;
> > B: r2 = READ_ONCE(y);
> >
> > Should B be ordered after A? I don't see how any CPU could hope to
> > excute B before A, but maybe I'm missing something.
> >
>
> Agreed.
>
> > There's another twist, connected with the fact that herd7 can't detect
> > control dependencies caused by unexecuted code. If we have:
> >
> > A: r1 = READ_ONCE(x);
> > if (r1)
> > WRITE_ONCE(y, 5);
> > r2 = READ_ONCE(y);
> > B: WRITE_ONCE(z, r2);
> >
> > then in executions where x == 0, herd7 doesn't see any control
> > dependency. But CPUs do see control dependencies whenever there is a
> > conditional branch, whether the branch is taken or not, and so they will
> > never reorder B before A.
> >
>
> Right, because B in this example is a write, what if B is a read that
> depends on r2, like in my example? Let y be a pointer to a memory
> location, and initialized as a valid value (pointing to a valid memory
> location) you example changed to:
>
> A: r1 = READ_ONCE(x);
> if (r1)
> WRITE_ONCE(y, 5);
> C: r2 = READ_ONCE(y);
> B: r3 = READ_ONCE(*r2);
>
> , then A don't have the control dependency to B, because A and B is
> read+read. So B can be ordered before A, right?

Yes, I think that's right: Both C and B can be executed before A.

> > One last thing to think about: My original assessment or Björn's problem
> > wasn't right, because the dep in (dep ; rfi) doesn't include control
> > dependencies. Only data and address. So I believe that the LKMM
>
> Ah, right. I was mising that part (ctrl is not in dep). So I guess my
> example is pointless for the question we are discussing here ;-(
>
> > wouldn't consider A to be ordered before B in this example even if x
> > was nonzero.
>
> Yes, and similar to my example (changing B to a read).
>
> I did try to run my example with herd, and got confused no matter I make
> dep; [Plain]; rfi as to-r (I got the same result telling me a reorder
> can happen). Now the reason is clear, because this is a ctrl; rfi not a
> dep; rfi.
>
> Thanks so much for walking with me on this ;-)

You're welcome. At this point, it looks like the only remaining
question is whether to include (dep ; [Plain] ; rfi) in to-r. This
doesn't seem to be an urgent question.

Alan