How to properly use std::enable_if on a constructor

Question

This question combines several pieces of code and is a bit complicated, but I tried slimming it down as much as possible.

I am trying to use std::enable_if

Answer 1

Dependent names

typename function_traits<Lambda>::template arg<0>::type
                                  ^^^^^^^^

See this post for more information on dependent names and when template or typename is needed.

enable_if

typename = std::enable_if_t<condition>

should instead be

std::enable_if_t<condition>* = nullptr

as @Jarod42 mentioned. This is because the constructors would otherwise be identical and unable to be overloaded. That their default values differ doesn't change this fact. See this for more information.

Putting it together is

template<typename Lambda, std::enable_if_t<std::is_same_v<typename function_traits<Lambda>::template arg<0>::type, a_type>>* = nullptr>
A(const Lambda&);

Live

Side note

function_traits won't work with either overloaded or templated operator(), it can instead be replaced

template<typename T, typename... Args>
using return_type = decltype(std::declval<T>()(std::declval<Args>()...));

template<typename T, typename... Args>
using mfp = decltype(static_cast<return_type<T, Args...>(T::*)(Args...) const>(&T::operator()));

template<typename Lambda, mfp<Lambda, a_type> = nullptr>
A(const Lambda&);

To check if the callable can be called with the exact arguments without conversions.