Does it make sense to combine optional with reference_wrapper?

Question

It occurred to me that in C++ it is possible to use the type std::optional>. An object of this type is essentially a reference

Answer 1

Is there any conceptual difference between std::optional<std::reference_wrapper<T>> and T*?

std::optional<>, as the name already suggest, is meant to be used when we could have a value or might not have any value at all.

The equivalent of having no value for a T* object would be assigning nullptr to it, i.e.: the pointer will point to nowhere, as opposed to somewhere (or even anywhere, i.e.: uninitialized). It can be said that std::optional<> exports the concept of nullptr for pointers to any arbitrary type. So, I would say they are conceptually very similar, being the std::option<> approach a kind of generalization.

Is there any practical difference? Are there situations where it might be advisable to choose std::optional<std::reference_wrapper<T>> over T*?

I can think of the size. std::optional<> contains an internal flag for indicating the presence/absence of a value, whereas for T* the nullptr is encoded directly as one of the values the pointer can store. So a std::optional<std::reference_wrapper<T>> object will be larger than a T*.

When it comes to safety, unlike T*, std::optional<> provides the member function value() which throws an exception if there is no value (it provides as well as the unsafe operator*() as T* does).

Also, using std::optional<std::reference_wrapper<T>> instead of T* , for example, as a function's return value may indicate in a more explicit way that there might be no value at all.

Answer 2

The main difference between std::optional<std::reference_wrapper<T>> and T* is that with T* you have to think about who owns the memory that is pointed to.

If a function returns T* you have to know if you are responsible for freeing the memory or someone else is. That's not something you have to be concerned with when it's a reference.