Is it possible to pass a concept as a template parameter?

user_1953032 0
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Is it possible to pass a concept as a template parameter? For example:

I want to do something like this:

template <typename t, typename u> concept range_of =
    range<t> &&
    requires (t a) {
        {*a.begin()} -> std::same_as<u &>;
    };

But instead of giving the exact type u I want to give it a concept:

template <typename t, {{concept u}}> concept constrained_range =
    range<t> &&
    requires (t a) {
        {*a.begin()} -> u;
    };
c++-concepts c++
2 Answers

It's possible to cheat a bit since we can pass lambda functions as template arguments ; a concept is just a meta-function from the domain of types to the boolean domain after all.

Here is an example:

template<auto F, typename T>
using check = std::conditional_t<
      F.template operator()<T>()
    , std::true_type
    , std::false_type>;

#define CONCEPT(TheConcept) \
  [] <typename T> () consteval { return TheConcept<T>; }

static_assert(check<CONCEPT(std::floating_point), float>::value);

Demo on Godbolt: https://gcc.godbolt.org/z/fGf45nqTc

user_14956270

Currently it's not possible to pass a concept as a template parameter, but you can get around it by passing a template template:

#include <optional>

template <typename T>
concept Optional = requires {
  typename T::value_type;
  // ...
};


template <template<typename> typename Q, typename T>
concept OptionalOf = Optional<T> && Q<typename T::value_type>::value;

which you can use like this (on compiler-explorer):

constexpr bool is_optional_of_int = OptionalOf<std::is_integral, std::optional<int>>;

or this:

template <typename T>
struct constrained_range {
  static constexpr bool value = range<T> && requires (T a) {
        {*a.begin()} -> u;
    };
};

constexpr bool is_optional_of_constrained_range = OptionalOf<constrained_range, std::optional<int>>;
user_49874700
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