Memory model release/acquire mode interactions of relaxed atomic operations

[Jonathan Wakely]

On 4 May 2017 at 13:40, Andrew Haley wrote:
> On 04/05/17 13:11, Toebs Douglass wrote:
>> On 04/05/17 14:03, Andrew Haley wrote:
>>> On 04/05/17 12:46, Toebs Douglass wrote:
>>>> Inherently volatile, kindafing?  lay off the optimisation and caching in
>>>> registers, etc?
>>>
>>> Much stronger than volatile.
>>>
>>> C++ atomic types are sequentially consistent by default.  Every
>>> processor sees memory as if the atomic loads and stores of all the
>>> processors had been executed in some sequential order.  Every
>>> processor sees that same total ordering.  GCC will throw in as many
>>> barrier instructions as are necessary to make that work.
>>
>> This being true if and only if the atomic load/store functions are used,
>> right?
>
> It works for all accesses: atomicity is part of the type.

Although there are also relaxed atomic operations (on atomic types)
which are not synchronization operations. Not all atomic operations
provide sequential consistency.



>> and I suspect those functions are only issuing memory barriers?
>
> And loads/stores, of course.
>
>> they do not use atomic operations themselves.
>
> Umm, what?  How can an atomic operation not use an atomic operation?
>
>> If this is so, then all this is true, but there is within this
>> absolutely no guarantee than any store on any core will ever
>> *actually be seen* by any other core.  If they *are* seen, *they
>> will be in the correct order*, but there is no guarantee they *will*
>> be seen.
>
> It's only about ordering.  However, you can guarantee that if some
> processor writes to a (SC) atomic variable, then that processor will
> have observed all writes before then in the total order.
>
>> I expect you know this, Andrew, but I suspect it's not commonly
>> understood, so it's useful to state it explicitly for the record and
>> possibly for the OP.  Also, I may be wrong, and if I am, I'd like to
>> find out :-)
>
> See above.  If you take part in the total order by doing an atomic
> store, all of everyone else's atomic writes are visible to you before
> that store.  This must be true, by the definition of total order.
>
> Andrew.