我想在编译时检查是否为类型_name
和参数Ret
定义了用户文字Arg
。虽然我有半解决方案,但它要求文字operator
至少定义一次:
#include <iostream>
#include <type_traits>
struct one { };
struct two { };
// we need at least one of these definitions for template below to compile
one operator"" _x(char const*) {return {};}
two operator"" _x(unsigned long long int) {return {};}
template<class T, class S, class = void>
struct has_literal_x : std::false_type
{ };
template<class T, class S>
struct has_literal_x <T, S,
std::void_t<decltype((T(*)(S))(operator"" _x))>
> : std::true_type
{ };
int main()
{
std::cout << has_literal_x<one, char const*>::value << std::endl;
std::cout << has_literal_x<two, unsigned long long int>::value << std::endl;
std::cout << has_literal_x<one, unsigned long long int>::value << std::endl;
std::cout << has_literal_x<two, char const*>::value << std::endl;
std::cout << has_literal_x<int, char const*>::value << std::endl;
}
输出:
1
1
0
0
0
但是,如果没有至少一个可能重载的用户文字的定义,则此解决方案将不起作用。是否有任何方法可以检查它甚至是不存在的文字(可能我们可以检查类X
是否有成员member
,但我不知道它是否已经知道)在这种情况下可行)?
答案 0 :(得分:9)
是否可以检查是否为给定的类型和参数定义了用户文字?
(简短)答案是是。
例如,您可以在示例代码中使用以下特化:
template<class T, class S>
struct has_literal_x <T, S,
std::enable_if_t<std::is_same<decltype(operator""_x(std::declval<S>())), T>::value>
> : std::true_type
{ };
很快就会变成:
#include <iostream>
#include <type_traits>
#include <utility>
struct one { };
struct two { };
//one operator"" _x(char const*) { return {}; }
//two operator"" _x(unsigned long long int) { return {}; }
template<class T, class S, class = void>
struct has_literal_x : std::false_type
{ };
template<class T, class S>
struct has_literal_x <T, S,
std::enable_if_t<std::is_same<decltype(operator""_x(std::declval<S>())), T>::value>
> : std::true_type
{ };
int main()
{
std::cout << has_literal_x<one, char const*>::value << std::endl;
std::cout << has_literal_x<two, unsigned long long int>::value << std::endl;
std::cout << has_literal_x<one, unsigned long long int>::value << std::endl;
std::cout << has_literal_x<two, char const*>::value << std::endl;
std::cout << has_literal_x<int, char const*>::value << std::endl;
}
输出是预期的:0
所有输出。
另一种在C ++ 14中做到这一点的方法(主要受@ Jarod42的this答案启发)是通过模板变量。
举个例子:
template<typename T, typename S, typename = void>
constexpr bool has_literal_v = false;
template<typename T, typename S>
constexpr bool has_literal_v<T, S, std::enable_if_t<std::is_same<decltype(operator""_x(std::declval<S>())), T>::value>> = true;
main
会改为:
int main()
{
std::cout << has_literal_v<one, char const*> << std::endl;
std::cout << has_literal_v<two, unsigned long long int> << std::endl;
std::cout << has_literal_v<one, unsigned long long int> << std::endl;
std::cout << has_literal_v<two, char const*> << std::endl;
std::cout << has_literal_v<int, char const*> << std::endl;
}
我觉得它很容易阅读,而且是constexpr
变量。还有什么?
答案 1 :(得分:5)
使用is_detected
函数系列,您可以执行
template <typename T>
using has_literal_x_type = decltype(operator"" _x(std::declval<T>()));
template <typename Ret, typename T>
using has_literal_x = std::is_same<Ret, detected_t<has_literal_x_type, T>>;
用
进行测试static_assert(!has_literal_x<one, char const*>::value, "unexpected");
static_assert(!has_literal_x<one, unsigned long long int>::value, "unexpected");
static_assert(!has_literal_x<two, char const*>::value, "unexpected");
static_assert(!has_literal_x<two, unsigned long long int>::value, "unexpected");
static_assert(!has_literal_x<int, char const*>::value, "unexpected");