Must I call atomic load/store explicitly?

Question

C++11 introduced the std::atomic<> template library. The standard specifies the store() and load() operations to atomically set / get a varia

Answer 1

Are assignment and access operations for non-reference types also atomic?

Yes, they are. atomic<T>::operator T and atomic<T>::operator= are equivalent to atomic<T>::load and atomic<T>::store respectively. All the operators are implemented in the atomic class such that they will use atomic operations as you would expect.

I'm not sure what you mean about "non-reference" types? Not sure how reference types are relevant here.

Answer 2

You can do both, but the advantage of load()/store() is that they allow to specify memory order. It is important sometimes for performance, where you can specify std::memory_order_relaxed while atomic<T>::operator T and atomic<T>::operator= would use the most safe and slow std::memory_order_seq_cst. Sometimes it is important for correctness and readability of your code: although the default std::memory_order_seq_cst is most safe thus most likely to be correct, it is not immediately clear for the reader what kind of operation (acquire/release/consume) you are doing, or whether you are doing such operation at all (to answer: isn't relaxed order sufficient here?).