将指针和const添加到std :: tuple <Types…>

Question

I'm trying to achieve the following using the magic of C++11 templates:

Suppose I have a type like this:

using my_types = std::tuple<char, int, float>;

Having this, I'd like to get a tuple of pointers to both const and not values, ie:

std::tuple<char *, int *, float *, const char *, const int *, const float *>;

My solution for now:

template<typename T>
struct include_const {};

template<typename... Types>
struct include_const<std::tuple<Types...>> {
  using type = std::tuple<Types..., typename std::add_const<Types>::type...>;
};

This gives std::tuple<types, const types> . To get pointers, I can use:

template<typename T>
struct add_ptr {};

template<typename... Types>
struct add_ptr<std::tuple<Types...>> {
  using type = std::tuple<typename std::add_pointer<Types>::type...>;
};

This works, but I would like this to get a little more general: I want to have a template<trait, Types...> add_ptr that gives me pointers to both Types... and trait<Types>::type... , so the usage could be the following:

add_ptr<std::add_const, my_types> is the tuple i mentioned before add_ptr<std::add_volatile, my_types> gives std::tuple<char *, volatile char *, ...>

I would appreciate some hints on how this can be achieved. I'm not yet a template magician and would appreciate some help

Answer 1

Use a template template-parameter

template<template<typename> class Trait, typename U>
struct add_ptr {};

template<template<typename> class Trait, typename... Types>
struct add_ptr<Trait, std::tuple<Types...>> {
  using type = std::tuple<
                    typename std::add_pointer<Types>::type...,
                    typename std::add_pointer<
                        typename Trait<Types>::type
                    >::type...
                >;
};

Then

add_ptr<std::add_const, my_types>::type

will be

std::tuple<char *, int *, float *, char const *, int const *, float const *>

Live demo