Checking type of parameter pack using enable_if

Question

Since there is a restriction on allowed non-type variadic templates, I am trying to write a function taking an arbitrary number of doubles using enable_if. In essence, I want to do something like:

    template<typename... T,
    typename = typename std::enable_if<std::is_convertible<T, double>::value, T>::type>
    foo(T... t){ /* code here */ }

I'm opting to put the enable_if as a default value for an unnamed parameter since my function is actually a constructor and will not have a return value. This would work for a single parameter, but as it's a variadic template T is a parameter pack, and the above code is not valid. So, how can I check every parameter is convertible to a double?

Answer 1

The bool_pack trick again.

template<bool...> struct bool_pack;
template<bool... bs> 
using all_true = std::is_same<bool_pack<bs..., true>, bool_pack<true, bs...>>;

Then

template<class R, class... Ts>
using are_all_convertible = all_true<std::is_convertible<Ts, R>::value...>;

and finally

template<typename... T,
typename = typename enable_if<are_all_convertible<double, T...>::value>::type>
foo(T... t){ /* code here */}

Answer 2

You could use fold expression in c++17 to do the same thing as other answers posted here but without the hassle of creating templates.

#include <type_traits>

template <typename... T, typename = 
    typename std::enable_if<
        (true && ... && std::is_convertible_v<T, ___YOUR_TYPE___>),
        void
    >::type
>
constexpr auto foo(T...) noexcept {
        // your code 
}