Why is std::nullopt_t part of the C++ standard?

问题

I don't understand the reasoning for the inclusion of std::nullopt_t in the standard. Does it exist strictly for convenience, or is it required in some niche circumstances?

To be clear, I understand that it is used as an argument to construct empty std::optional objects. But considering a default constructor for std::optional already exists, there seems to be no obvious motivation for the existence of std::nullopt_t. Must such a constructor and assignment operator exist for std::optional to conform to a particular concept? If so, which concept?

回答1:

nullopt_t is the type of nullopt which indicates disengaged optional state. nullopt allows disambiguating overloads such as (example from the optional proposal):

void run(complex<double> v);
void run(optional<string> v);

run(nullopt);              // pick the second overload
run({});                   // ambiguous

回答2:

C++ reference says it all:

std::nullopt_t is an empty class type used to indicate optional type with uninitialized state. In particular, std::optional has a constructor with nullopt_t as a single argument, which creates an optional that does not contain a value.

std::nullopt_t

来源：https://stackoverflow.com/questions/49617785/why-is-stdnullopt-t-part-of-the-c-standard