Question

from c++11 29.3-p3:

There shall be a single total order S on all memory_order_seq_cst operations, consistent with the "happens before" order and modification orders for all affected locations, such that each memory_order_seq_cst operation B that loads a value from an atomic object M observes either one of the following values:

— the result of the last modification A of M that precedes B in S, if it exists, or

— if A exists, the result of some modification of M in the visible sequence of side effects with respect to B that is not memory_order_seq_cst and that does not happen before A, or

— if A does not exist, the result of some modification of M in the visible sequence of side effects with respect to B that is not memory_order_seq_cst.

[ Note: Although it is not explicitly required that S include locks, it can always be extended to an order that does include lock and unlock operations, since the ordering between those is already included in the "happens before" ordering. –end note ]

In the last note, what does it mean by "always"?
I can understand that any certain implementation can be designed to support such an extended S. But in some general implementation that wasn’t designed for it, I don’t see that S can be extended with the described properties.

Does it mean an extended order that also satisfies the same visibility properties here?

I had sent this question to comp.std.c++ but got no answers there. http://groups.google.com/group/comp.std.c++/browse_frm/thread/5242fa70d0594d1b#

Answer 1

What does it mean by “always”?

Always means that there exists a total order S+l on seq_cst ops U lock ops that is consistent with happens-before and S.

Why is that

Fact 1: S is a total order consistent with HB.

Fact 2: Lock operations are ordered in the HB partial order, because they are acquire/release operations.

Fact 3: There are no cycles in HB.

Lemma 1: There are no cycles in S' = S union HB.

Proof: If there were any cycles in S', they would be in the form Aop1 <S Aop2 <HB Aop1, because any two atomic operations are comparable in S and happens-before is transitive. That would contradict with Fact 1. QED

Conclusion: Because for each partial order (= without cycles) there exists its extension to total order (cf. topological sorting), there is a total order extending S'. So you just pick atomics and lock operations from it and get S+l.

But in some general implementation that wasn’t designed for it, I don’t see that S can be extended so.

Such an implementation would not satisfy mem_order_seq_cst requirements.