c++ How to get arity of a constructor?

Question

Given a template parameter class T that has a single constructor (no copy or move constructor either) and zero default arguments, is there some way to find the arity of T(...)?

My attempts so far

#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;
}

@Jarod No, it and the move constructor are marked as deleted. There is only **one** constructor available. — Brian Rodriguez, Sep 16 '15 at 16:44
@BrianRodriguez: I mean that deleted functions are part of overload [Demo](http://coliru.stacked-crooked.com/a/2bae194578965da1). — Jarod42, Sep 16 '15 at 16:53
@Columbo Well, the types should _theoretically_ be known statically (That's how I'll be using this). Do you mean I'd have to explicitly list the types when I try to deduce them? That'd kind of defeat my goal here... — Brian Rodriguez, Sep 16 '15 at 16:58

Columbo · Accepted Answer · 2015-09-17T07:05:45.823

If we make the following assumptions:

The parameters are either fixed types or (entirely unrestricted!) catch-all parameters (not e.g. std::basic_string<CharT>)
The parameters are MoveConstructible

, then

#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>;

…should work as expected. Demo.
Note that the above approach is easily translatable into C++11.

c++ How to get arity of a constructor?

1 Answers1