Re: XDP socket rings, and LKMM litmus tests

On Thu, Mar 04, 2021 at 09:26:31AM +0800, Boqun Feng wrote:

On Wed, Mar 03, 2021 at 03:22:46PM -0500, Alan Stern wrote:

quoted

Which brings us back to the case of the

	dep ; rfi

dependency relation, where the accesses in the middle are plain and 
non-racy.  Should the LKMM be changed to allow this?

For this particular question, do we need to consider code as the follow?

	r1 = READ_ONCE(x);  // f
	if (r == 1) {
		local_v = &y; // g
		do_something_a();
	}
	else {
		local_v = &y;
		do_something_b();
	}

	r2 = READ_ONCE(*local_v); // e

, do we have the guarantee that the first READ_ONCE() happens before the
second one? Can compiler optimize the code as:

	r2 = READ_ONCE(y);
	r1 = READ_ONCE(x);

Well, it can't do that because the compiler isn't allowed to reorder
volatile accesses (which includes READ_ONCE).  But the compiler could
do:

	r1 = READ_ONCE(x);
	r2 = READ_ONCE(y);

	if (r == 1) {
		do_something_a();
	}
	else {
		do_something_b();
	}

? Although we have:

	f ->dep g ->rfi ->addr e

This would be an example of a problem Paul has described on several
occasions, where both arms of an "if" statement store the same value
(in this case to local_v).  This problem arises even when local
variables are not involved.  For example:

	if (READ_ONCE(x) == 0) {
		WRITE_ONCE(y, 1);
		do_a();
	} else {
		WRITE_ONCE(y, 1);
		do_b();
	}

The compiler can change this to:

	r = READ_ONCE(x);
	WRITE_ONCE(y, 1);
	if (r == 0)
		do_a();
	else
		do_b();

thus allowing the marked accesses to be reordered by the CPU and
breaking the apparent control dependency.

So the answer to your question is: No, we don't have this guarantee,
but the reason is because of doing the same store in both arms, not
because of the use of local variables.

Alan