Compare and swap C++0x
The answer here is not trivial. Exactly what happens and what is meant is dependent on many things. For basic understanding of cache coherence/memory perhaps my recent blog entries might be helpful:
- CPU Reordering – What is actually being reordered?
- CPU Memory – Why do I need a mutex?
But that aside, let me try to answer a few questions. First off the below function is being very hopeful as to what is supported: very fine-grained control over exactly how strong a memory-order guarantee you get. That's reasonable for compile-time reordering but often not for runtime barriers.
compare_swap( C& expected, C desired,
memory_order success, memory_order failure )
Architectures won't all be able to implement this exactly as you requested; many will have to strengthen it to something strong enough that they can implement. When you specify memory_order you are specifying how reordering may work. To use Intel's terms you will be specifying what type of fence you want, there are three of them, the full fence, load fence, and store fence. (But on x86, load fence and store fence are only useful with weakly-ordered instructions like NT stores; atomics don't use them. Regular load/store give you everything except that stores can appear after later loads.) Just because you want a particular fence on that operation won't mean it is supported, in which I'd hope it always falls back to a full fence. (See Preshing's article on memory barriers)
An x86 (including x64) compiler will likely use the LOCK CMPXCHG instruction to implement the CAS, regardless of memory ordering. This implies a full barrier; x86 doesn't have a way to make a read-modify-write operation atomic without a lock prefix, which is also a full barrier. Pure-store and pure-load can be atomic "on their own", with many ISAs needing barriers for anything above mo_relaxed, but x86 does acq_rel "for free" in asm.
This instruction is lock-free, although all cores trying to CAS the same location will contend for access to it so you could argue it's not really wait-free. (Algorithms that use it might not be lock-free, but the operation itself is wait-free, see wikipedia's non-blocking algorithm article). On non-x86 with LL/SC instead of locked instructions, C++11 compare_exchange_weak is normally wait-free but compare_exchange_strong requires a retry loop in case of spurious failure.
Now that C++11 has existed for years, you can look at the asm output for various architectures on the Godbolt compiler explorer.
In terms of memory sync you need to understand how cache-coherence works (my blog may help a bit). New CPUs use a ccNUMA architecture (previously SMP). Essentially the "view" on the memory never gets out-of-sync. The fences used in the code don't actually force any flushing of cache to happen per-se, only of the store buffer committing in flight stores to cache before later loads.
If two cores both have the same memory location cached in a cache-line, a store by one core will get exclusive ownership of the cache line (invalidating all other copies) and marking its own as dirty. A very simple explanation for a very complex process
To answer your last question you should always use the memory semantics that you logically need to be correct. Most architectures won't support all the combinations you use in your program. However, in many cases you'll get great optimizations, especially in cases where the order you requested is guaranteed without a fence (which is quite common).
-- Answers to some comments:
You have to distinguish between what it means to execute a write instruction and write to a memory location. This is what I attempt to explain in my blog post. By the time the "0" is committed to 0x100, all cores see that zero. Writing integers is also atomic, that is even without a lock, when you write to a location all cores will immediately have that value if they wish to use it.
The trouble is that to use the value you have likely loaded it into a register first, any changes to the location after that obviously won't touch the register. This is why one needs mutexes or atomic<T> despite a cache coherent memory: the compiler is allowed to keep plain variable values in private registers. (In C++11, that's because a data-race on non-atomic variables is Undefined Behaviour.)
As to contradictory claims, generally you'll see all sorts of claims. Whether they are contradictory comes right down to exactly what "see" "load" "execute" mean in the context. If you write "1" to 0x100, does that mean you executed the write instruction or did the CPU actually commit that value. The difference created by the store buffer is one major cause of reordering (the only one x86 allows). The CPU can delay writing the "1", but you can be sure that the moment it does finally commit that "1" all cores see it. The fences control this ordering by making the thread wait until a store commits before doing later operations.