`std::function` returning a `concept`

Question

Proceeding with my studies on concepts, and after looking here, here and here, among other pages, I wonder how could I declare a function type that returns a std::optional of another concept, like below:

template <typename t>
concept event = requires(t p_t) {
    { t::id } -> std::same_as<const size_t &>;
    requires std::default_initializable<t>;
};

using func = std::function<std::optional<event>(int)>;

gcc (Ubuntu 11.3.0-1ubuntu1~22.04) 11.3.0 reports Too few template arguments for concept 'event'

I tried

template <typename t>
using func = std::function<std::optional<event<t>>(int)>;

which generated the error Template argument for template type parameter must be a type

Is there a way to do it?

Answer 1

That is not possible. A concept is not a type. It's purpose is to make setting constraints easier (for example in contexts where you'd use std::enable_if pre C++20).

Example:

// Pre C++20 SFINAE
template<typename T>
struct event_sfinae {
    static constexpr bool value = ...;
};

template<typename T, typename = std::enable_if_t<event_sfinae<T>::value>>
void foo1(T arg) {
    ...
}

// C++20 concept version
template<typename T>
concept event_concept = ...; // Define concept

template<event_concept T> // only types that satisfy the constraints of 'event_concept' allowed
void foo2(T arg) {
    ...
}