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.
