A std::reference_wrapper is guaranteed to be initialized (but can be rebound, unlike a normal reference).
Being a wrapper class, what's the reason for std::reference_wrapper not to overload operator-> (and operator*)?
Instead, it is some other functions, like the get() method, to 'unwrap'.
I'm interested to hear what the deeper language reason is to why it is so (canonical or not).
std::reference_wrapper is designed to act like a C++ reference, not a pointer.
The operator-> is conventionally overloaded by types that behave like pointers (e.g., smart pointers such as std::unique_ptr or std::shared_ptr). Overloading operator-> for std::reference_wrapper would make it behave more like a pointer, which is not its intended semantic.
std::optional uses these operators because it deals with the potential presence of a value, similar to how a pointer might be "null" or point to a value. Whereas std::reference_wrapper aims to mimic a true reference as closely as possible, and true references use the dot (.) operator for member access. It must always refer to a valid object. It cannot be "empty" or "null" in the way std::optional can.
->or*on a normal reference. Why wouldreference_wrapperallow it?operator->andoperator*was discussed at std-proposals last year.