alias for nested template type

Question

I have the following code:

template <template <typename, typename...> typename trait_t, typename arg_t>
struct BindFirst
{
    template <typename... arg_ts>
    using result_t = trait_t<arg_t, arg_ts...>;
};

#define BIND_FIRST(trait_t, arg_t) BindFirst<trait_t, arg_t>::template result_t

you can use it to bind the first argument of a trait like this:

BIND_FIRST(std::is_same, double)

The result is equivalent to:

template <typename T>
struct IsInt : std::is_same<double, T> { };

The difference is, that you can use it inline. For example like this:

using result_t = find_t<type_list, BIND_FIRST(std::is_same, double)>;

This works but i like to avoid the define. I tried to use an alias. But I have no idea how to apply it. Is there any way to replace the define?

Answer 1

You can use templates with using to create an alias template, just like you did for result_t .

template <typename... Args>
using IsDouble = BindFirst<std::is_same, double>::template result_t<Args...>;

You can limit the Args... to a single type T as well, it doesn't have to be variadic.

Edit: If your goal is to reduce boilerplate, you may want to opt for something like this

template <typename T, typename U>
using IsSameAs = std::is_same<T, U>;

template <typename T>
using IsDouble = IsSameAs<double, T>;