c ++ 11原子排序：锁的扩展总命令memory_order_seq_cst

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.