c++如何获得构造函数的数量？

Question

给定一个模板参数类 T，它有一个单个 构造函数(也没有复制或移动构造函数)和零 默认参数，有没有办法找到 T(...) 的元数？

到目前为止我的尝试

#include <iostream>
#include <string>
#include <vector>

template <typename F> struct function_arity;

template <typename R, typename... Args>
struct function_arity<R (Args...)>
    : std::integral_constant<std::size_t, sizeof...(Args)> {};

template <typename R, typename... Args>
struct function_arity<R (*)(Args...)> : function_arity<R (Args...)> {};

template <typename R, typename... Args>
struct function_arity<R (&)(Args...)> : function_arity<R (Args...)> {};

template <typename R, typename C, typename... Args>
struct function_arity<R (C::*)(Args...) const> : function_arity<R (Args...)> {};

template <typename R, typename C, typename... Args>
struct function_arity<R (C::*)(Args...)> : function_arity<R (Args...)> {};

template <typename C>
struct function_arity : function_arity<decltype(&C::operator())> {};

struct no_copy { no_copy() = default; no_copy(const no_copy&) = delete; };
struct no_move { no_move() = default; no_move(no_move&&) = delete; };

struct A : no_copy, no_move { A(int, float) { std::cout << "A!\n"; }; };
struct B : no_copy, no_move { B(double) { std::cout << "B!\n"; }; };
struct C : no_copy, no_move { C() { std::cout << "C!\n"; }; };

int main()
{
    std::cout << function_arity<&A::A>::value << "\n";
    return 0;
}

Answer 1

如果我们做出如下假设:

• 参数要么是固定类型，要么(完全不受限制!)包罗万象的参数(不是例如 std::basic_string<CharT> )
• 参数是MoveConstructible

然后

#include <type_traits>
#include <utility>

namespace detail {
    template <typename Ignore>
    struct anything {
        template <typename T,
                  typename=std::enable_if_t<not std::is_same<Ignore, std::decay_t<T>>{}>>
        operator T&&();
    };

    template <typename U, typename=void, typename... args>
    struct test : test<U, void, args..., anything<U>> {};
    template <typename U, typename... args>
    struct test<U, std::enable_if_t<std::is_constructible<U, args...>{}
                                 && sizeof...(args) < 32>, args...>
        : std::integral_constant<std::size_t, sizeof...(args)> {};
    template <typename U, typename... args>
    struct test<U, std::enable_if_t<sizeof...(args) == 32>, args...>
        : std::integral_constant<std::size_t, (std::size_t)-1> {};
}

template <typename U>
using ctor_arity = detail::test<U, void>;

...应该按预期工作。 Demo .
请注意，上述方法很容易转换为 C++11。