如何在具有相同签名的两个构造函数之间进行有条件的切换？

Question

Given a class with template parameters typename T and class Tuple I want to provide a special constructor, if Tuple has std::vector -like member functions reserve and push_back . If Tuple has no such member functions, then I want to provide a special constructor, if Tuple is constructible from variadic arguments of types convertible to T , ie

template<typename T, class Tuple>
class vector
{
    template<typename... Elements,
        typename = decltype(std::declval<Tuple>().push_back(T())),
        typename = decltype(std::declval<Tuple>().reserve(size_type()))>
    vector(Elements&&... elements)
    { /* ... */ }

    template<typename... Elements, typename = typename = decltype(Tuple{ static_cast<T>(std::declval<Elements>())... })>
    vector(Elements&&... elements)
    { /* ... */ }
};

Question 1: Clearly, in the code above the compiler doesn't know that I want to take the first constructor whenever possible. How can I achieve the desired behavior anyway?

Question 2: Assuming that the first constructor doesn't exist, why does the following code lead to the compiler error "cannot convert from initializer list to vector<double, Tuple<double>> ":

template<typename T>
class Tuple
{
public:
    Tuple() { }
    Tuple(std::initializer_list<T>) { }
};

int main()
{
    vector<double, Tuple<double>> x = { 1, 2, 3 };
    return 0;
}

Answer 1

Let's say we have two type traits:

template <class T, Tuple> struct satisfies_A;
template <class T, Tuple> struct satisfies_B;

And we want to provide a constructor to call one or the other based on satisfaction. We can first add direct constructors for each case with some additional tag type:

template <class> struct tag{};

template <typename T, class Tuple>
class vector
{
    struct A_tag { };
    struct B_tag { };
    struct na_tag { }; // if you want to support a fallback?

public:
    template <class U=T, class UTuple=Tuple, class... Elements,
        class = std::enable_if_t<satsfies_A<U,UTuple>::value>>
    vector(tag<A_tag>, Elements&&... );

    template <class U=T, class UTuple=Tuple, class... Elements,
        class = std::enable_if_t<satsfies_B<U,UTuple>::value>>
    vector(tag<B_tag>, Elements&&... );
};

Those constructors do whatever it is you want them to do based on the two different type traits. Now, we can introduce a type like:

using ctor_tag = std::conditional_t<
    satisfies_A<T, Tuple>::value,
    A_tag,
    std::conditional_t<
        satisfies_B<T, Tuple>::value,
        B_tag,
        na_tag>>; // or just void

And forward as appropriate:

template <class Element, class... Elements,
    class = std::enable_if_t<!is_template<tag, std::decay_t<Element>>::value>,
    class = std::enable_if_t<std::is_constructible<vector, tag<ctor_tag>, Element&&, Elements&&...>::value>>
vector(Element&& element, Elements&&... elements)
: vector(tag<ctor_tag>{}, std::forward<Element>(element), std::forward<Elements>(elements)...)
{ }

Something like that.

Answer 2

It's unclear to me what behavior you want when the template and constructor arguments are neither vector-like nor tuple-like. Assuming you simply want these two constructors to be ineligible and don't want some default/fallback behavior when attempting to call them (ie, you'll provide other constructors that will be called instead), here's an attempt at solving your specific problem:

#include <type_traits>
#include <utility>

namespace detail {

template<typename...>
struct void_t_helper { using type = void; };

template<typename... Ts>
using void_t = typename void_t_helper<Ts...>::type;

template<typename ElemT, typename T, typename SizeT = typename T::size_type>
auto is_vectorlike(int, T& t)
 -> decltype(
        t.push_back(std::declval<ElemT>()),
        void(t.reserve(std::declval<SizeT>())),
        std::true_type()
    );

template<typename, typename T>
std::false_type is_vectorlike(long, T&);

template<
    typename T, typename ElemT, typename... ArgTs,
    typename = void_t<decltype(static_cast<ElemT>(std::declval<ArgTs>()))...>
>
auto is_tuplelike(int)
 -> decltype(void(T{ std::declval<ArgTs>()... }), std::true_type());

template<typename...>
std::false_type is_tuplelike(long);

}

template<typename T, typename ElemT>
using is_vectorlike = decltype(detail::is_vectorlike<ElemT>(0, std::declval<T&>()));

template<typename T, typename ElemT, typename... ArgTs>
using is_tuplelike = decltype(detail::is_tuplelike<T, ElemT, ArgTs...>(0));

// ...

template<typename T, typename Tuple>
struct vector
{
    vector() = default;

    template<typename Element, typename... Elements, typename = std::enable_if_t<
        is_vectorlike<Tuple, T>{} || is_tuplelike<Tuple, T, Element&&, Elements&&...>{}
    >>
    explicit vector(Element&& e, Elements&&... es)
      : vector(
            is_vectorlike<Tuple, T>{},
            is_tuplelike<Tuple, T, Element&&, Elements&&...>{},
            std::forward<Element>(e),
            std::forward<Elements>(es)...
        )
    { }

private:
    template<typename _, typename... Elements>
    vector(std::true_type, _, Elements&&...)
    { /*vector-like*/ }

    template<typename... Elements>
    vector(std::false_type, std::true_type, Elements&&...)
    { /*tuple-like*/ }
};

Nb it seems to me that you'd want to make sure all Elements are convertible to T for the vector-like case as well, but since you didn't say so, this doesn't. ;-]