On 4 May 2017 at 13:56, Toebs Douglass <toby@xxxxxxxxxxxxxx> wrote: > On 04/05/17 14:40, Andrew Haley wrote: >> On 04/05/17 13:11, Toebs Douglass wrote: > >>> 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. > > Why do the load()/store() functions exist? > > Like this, from the OP; > >> if (x.load(memory_order_acquire) == 10) >> assert (y == 20); > > Only to specify the memory model? Yes, to explicitly specify the memory ordering. x == 10 is exactly equivalent to x.load(memory_order_seq_cst) == 10. i.e. the default ordering is SC. > But if you access the variable without the function, how would the > compiler know what memory model you wanted to use? RTFM. http://en.cppreference.com/w/cpp/atomic/atomic/operator_T > Also, you are then saying that seemingly-unadorned accesses, i.e. > > x = 1; > > Will cause the compiler to emit not just compiler barriers, but memory > barriers? will the compiler in fact convert this into an atomic - in > the LL/SC sense - operation? (I don't think this is so?) x = 1 is exactly equivalent to x.store(memory_order_seq_cst). http://en.cppreference.com/w/cpp/atomic/atomic/operator%3D
