檢查2種類型是否具有可比性

Question

我有以下代碼來確定2種類型是否具有可比性。

template<typename T, typename U, typename = std::void_t<>> 
struct is_comparable 
    : std::false_type 
{}; 

template<typename T, typename U> 
struct is_comparable<T, U, std::void_t<decltype((std::declval<T>() == std::declval<U>()))>> 
    : std::true_type 
{}; 

這是一種可以接受的方式來實現我想要做的事嗎？你能看到這個設計有什麼問題嗎？

編輯

保持cdhowie的評論和亨利面機的記答案，這是現在的代碼的外觀。

namespace meta 
{ 
    template<typename T, typename U, typename = std::void_t<>> 
    struct has_equal_to_operator 
     : std::false_type 
    {}; 

    template<typename R, typename T, typename U, typename = std::void_t<>> 
    struct has_equal_to_operator_r 
     : std::false_type 
    {}; 

    template<typename T, typename U, typename = std::void_t<>> 
    struct has_nothrow_equal_to_operator 
     : std::false_type 
    {}; 

    template<typename R, typename T, typename U, typename = std::void_t<>> 
    struct has_nothrow_equal_to_operator_r 
     : std::false_type 
    {}; 

    template<typename T, typename U> 
    struct has_equal_to_operator<T, U, std::void_t<decltype(std::declval<T>() == std::declval<U>())>> 
     : std::true_type 
    {}; 

    template<typename R, typename T, typename U> 
    struct has_equal_to_operator_r<R, T, U, std::void_t<decltype(std::declval<T>() == std::declval<U>())>> 
     : std::is_convertible<decltype(std::declval<T>() == std::declval<U>()), R> 
    {}; 

    template<typename T, typename U> 
    struct has_nothrow_equal_to_operator<T, U, std::void_t<decltype(std::declval<T>() == std::declval<U>())>> 
     : std::bool_constant<noexcept(std::declval<T>() == std::declval<U>())> 
    {}; 

    template<typename R, typename T, typename U> 
    struct has_nothrow_equal_to_operator_r<R, T, U, std::void_t<decltype(std::declval<T>() == std::declval<U>())>> 
     : std::bool_constant<(noexcept(std::declval<T>() == std::declval<U>()) && std::is_convertible_v<decltype(std::declval<T>() == std::declval<U>()), R>)> 
    {}; 

    template<typename T, typename U> 
    inline constexpr auto has_equal_to_operator_v = has_equal_to_operator<T, U>::value; 

    template<typename R, typename T, typename U> 
    inline constexpr auto has_equal_to_operator_r_v = has_equal_to_operator_r<R, T, U>::value; 

    template<typename T, typename U> 
    inline constexpr auto has_nothrow_equal_to_operator_v = has_nothrow_equal_to_operator<T, U>::value; 

    template<typename R, typename T, typename U> 
    inline constexpr auto has_nothrow_equal_to_operator_r_v = has_nothrow_equal_to_operator_r<R, T, U>::value; 
} 

Answer 1

這裏是void_t解決像的問題。此外，我會檢查比較是否產生正確的類型（在這種情況下爲bool）。

#include <type_traits> 
#include <iostream> 

template < typename T, typename U > 
using equality_comparison_t = decltype(std::declval<T&>() == std::declval<U&>()); 

template < typename T, typename U, typename = std::void_t<> > 
struct is_equality_comparable 
    : std::false_type 
{}; 

template < typename T, typename U > 
struct is_equality_comparable < T, U, std::void_t< equality_comparison_t<T,U> > > 
    : std::is_same< equality_comparison_t<T,U>, bool > 
{}; 

struct X {}; 

struct Y { int operator==(Y const&) { return 1; } }; 

int main() 
{ 
    static_assert(false == is_equality_comparable<X, X>(), "!"); 
    static_assert(true == is_equality_comparable<std::string, std::string>(), "!!"); 
    static_assert(false == is_equality_comparable<int, std::string>(), "!!!"); 
    static_assert(true == is_equality_comparable<int, int>(), "!!!!"); 
    static_assert(false == is_equality_comparable<Y, Y>(), "!!!!!"); 
} 

Answer 2

你並不遠，但你並不需要C++ 17：這是不夠的C++ 11

#include <type_traits> 
#include <iostream> 

template <typename T, typename U, typename = void> 
struct is_comparable : std::false_type 
{}; 

template <typename T, typename U> 
struct is_comparable<T, U, 
    decltype((std::declval<T>() == std::declval<U>()), void())> 
    : std::true_type 
{}; 

struct X 
{ }; 

int main() 
{ 
    static_assert(false == is_comparable<X, X>(), "!"); 
    static_assert(true == is_comparable<std::string, std::string>(), "!!"); 
    static_assert(false == is_comparable<int, std::string>(), "!!!"); 
    static_assert(true == is_comparable<int, int>(), "!!!!"); 
}