为什么此C ++代码中的构造函数不明确，我该如何解决?

问题描述

在下面的代码中，编译器无法确定我要使用哪个构造函数.为什么，如何解决此问题? (在线示例)

In the below code, the compiler can't figure out which constructor I want to use. Why, and how do I fix this? (Live example)

#include <tuple>
#include <functional>
#include <iostream>

template<typename data_type, typename eval_type, typename Type1, typename Type2>
class A
{
public:
    using a_type = std::tuple<Type1, Type2>;
    using b_type = std::tuple<std::size_t,std::size_t>;

    inline explicit constexpr A(const std::function<data_type(a_type)>& Initializer,
        const std::function<eval_type(data_type)>& Evaluator,
        const Type1& elem1, const Type2& elem2)
    {
        std::cout << "idx_type" << std::endl;
    }
    inline explicit constexpr A(const std::function<data_type(b_type)>& Initializer,
        const std::function<eval_type(data_type)>& Evaluator,
        const Type1& elem1, const Type2& elem2)
    {
        std::cout << "point_type" << std::endl;
    }
};

int main()
{
    int a = 1;
    long long b = 2;
    auto c = A<double, double, long long, int>{
        [](std::tuple<long long,int> p)->double { return 1.0*std::get<0>(p) / std::get<1>(p); },
        [](double d)->double { return d; }, b,a
        };

    return 0;
}

推荐答案

如@SombreroChicken所述，std::function<R(Args...)>具有一个允许 any 可调用对象c对其进行初始化的构造函数，只要c(Args...)有效，并返回可转换为R的内容.

As @SombreroChicken mentioned, std::function<R(Args...)> has a constructor that allows any callable object c to initialize it, as long as c(Args...) is valid and returns something convertible to R.

要修复此问题，您可以使用一些SFINAE机械

To fix it, you may use some SFINAE machinery

#include <tuple>
#include <functional>
#include <iostream>
#include <type_traits>

template<typename data_type, typename Type1, typename Type2>
class A
{
    template<typename T>
    struct tag
    {
        operator T();
    };

public:
    using a_type = std::tuple<Type1, Type2>;
    using b_type = std::tuple<std::size_t,std::size_t>;

    template<typename C, std::enable_if_t<std::is_invocable_v<C, tag<b_type>>>* = nullptr>
    A(C&& initializer)
    {
        std::cout << "size_t" << std::endl;
    }

    template<typename C, std::enable_if_t<std::is_invocable_v<C, tag<a_type>>>* = nullptr>
    A(C&& initializer)
    {
        std::cout << "other" << std::endl;
    }
};

int main()
{
    auto c = A<double, long long, int>{
        [](std::tuple<long long, int> p) -> double { return 1; }
    };

    auto c2 = A<double, long long, int>{
        [](std::tuple<std::size_t, std::size_t>) -> double { return 2; }
    };
}

实时

在这里，如果可以分别用b_type或a_type调用可调用对象，则关闭构造函数.通过tag进行的额外间接操作可以禁止不同类型的元组之间的转换

Here, we turn off the constructor if the callable can be called with b_type or a_type respectively. The extra indirection through tag is there to disable the conversion between tuples of different types

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