C++11 Tagged Tuple

Question

I'm not aware of any existing class that does this, but it's fairly easy to throw something together using a std::tuple and an indexing typelist:

#include <tuple>
#include <iostream>

template<typename... Ts> struct typelist {
  template<typename T> using prepend = typelist<T, Ts...>;
};

template<typename T, typename... Ts> struct index;
template<typename T, typename... Ts> struct index<T, T, Ts...>:
  std::integral_constant<int, 0> {};
template<typename T, typename U, typename... Ts> struct index<T, U, Ts...>:
  std::integral_constant<int, index<T, Ts...>::value + 1> {};

template<int n, typename... Ts> struct nth_impl;
template<typename T, typename... Ts> struct nth_impl<0, T, Ts...> {
  using type = T; };
template<int n, typename T, typename... Ts> struct nth_impl<n, T, Ts...> {
  using type = typename nth_impl<n - 1, Ts...>::type; };
template<int n, typename... Ts> using nth = typename nth_impl<n, Ts...>::type;

template<int n, int m, typename... Ts> struct extract_impl;
template<int n, int m, typename T, typename... Ts>
struct extract_impl<n, m, T, Ts...>: extract_impl<n, m - 1, Ts...> {};
template<int n, typename T, typename... Ts>
struct extract_impl<n, 0, T, Ts...> { using types = typename
  extract_impl<n, n - 1, Ts...>::types::template prepend<T>; };
template<int n, int m> struct extract_impl<n, m> {
  using types = typelist<>; };
template<int n, int m, typename... Ts> using extract = typename
  extract_impl<n, m, Ts...>::types;

template<typename S, typename T> struct tt_impl;
template<typename... Ss, typename... Ts>
struct tt_impl<typelist<Ss...>, typelist<Ts...>>:
  public std::tuple<Ts...> {
  template<typename... Args> tt_impl(Args &&...args):
    std::tuple<Ts...>(std::forward<Args>(args)...) {}
  template<typename S> nth<index<S, Ss...>::value, Ts...> get() {
    return std::get<index<S, Ss...>::value>(*this); }
};
template<typename... Ts> struct tagged_tuple:
  tt_impl<extract<2, 0, Ts...>, extract<2, 1, Ts...>> {
  template<typename... Args> tagged_tuple(Args &&...args):
    tt_impl<extract<2, 0, Ts...>, extract<2, 1, Ts...>>(
      std::forward<Args>(args)...) {}
};

struct name {};
struct age {};
struct email {};

tagged_tuple<name, std::string, age, int, email, std::string> get_record() {
  return { "Bob", 32, "[email protected]"};
}

int main() {
  std::cout << "Age: " << get_record().get<age>() << std::endl;
}

You'll probably want to write const and rvalue get accessors on top of the existing one.

C++ does not have a struct type that can be iteratable like a tuple; it's either/or.

The closest you can get to that is through Boost.Fusion's struct adapter. This allows you to use a struct as a Fusion sequence. Of course, this also uses a series of macros, and it requires you to list the struct's members explicitly in the order you want to iterate over them. In the header (assuming you want to iterate over the struct in many translation units).

Actually my example is probably a bit unrealistic to implement. How about this?

You could implement something like that, but those identifiers need to actually be types or variables or something.

I have my own implementation to show off, wich can allow you not to declare the attributes on top of the file. A version with declared attributes exists too, but there is no need to define them, declaration is sufficient.

It is pure STL, of course, and do not use the preprocessor.

Example:

#include <named_tuples/tuple.hpp>
#include <string>
#include <iostream>
#include <vector>

namespace {
unsigned constexpr operator "" _h(const char* c,size_t) { return named_tuples::const_hash(c); }
template <unsigned Id> using at = named_tuples::attribute_init_int_placeholder<Id>;
using named_tuples::make_tuple;
}

int main() {
  auto test = make_tuple( 
      at<"nom"_h>() = std::string("Roger")
      , at<"age"_h>() = 47
      , at<"taille"_h>() = 1.92
      , at<"liste"_h>() = std::vector<int>({1,2,3})
      );

  std::cout 
    << test.at<"nom"_h>() << "\n"
    << test.at<"age"_h>() << "\n"
    << test.at<"taille"_h>() << "\n"
    << test.at<"liste"_h>().size() << std::endl;

  test.at<"nom"_h>() = "Marcel";
  ++test.get<1>();

  std::cout 
    << test.get<0>() << "\n"
    << test.get<1>() << "\n"
    << test.get<2>() << "\n"
    << test.get<3>().size() << std::endl;

  return 0;
}

Find the complete source here https://github.com/duckie/named_tuple. Feel free to read, it is quite simple.

The real problems you have to solve here are:

• Are the tags mandatory or optional?
• Are the tags unique? Is it enforced at compile time?
• In which scope does the tag reside? Your example seems to declare the tags inside the declaring scope instead of encapsulated in the type, which might not be optimal.

ecatmur proposed a good solution; but the tags are not encapsulated and the tag declaration is somehow clumsy. C++14 will introduce tuple addressing by type, which will simplify his design and guarantee uniqueness of the tags, but not solve their scope.

Boost Fusion Map can also be used for something similar, but again, declaring the tags is not ideal.

There is a proposal for something similar on the c++ Standard Proposal forum, which would simplify the syntax by associating a name to the template parameter directly.

This link lists different ways of implementing this (including ecatmur's solution) and presents a different use-case for this syntax.

Here's another way to do it, it's a bit uglier to define the types but it helps prevent errors at compile time because you define the pairs with a type_pair class (much like std::map). Adding a check to make sure your keys/name are unique at compile time is the next step

Usage:

   using user_t = tagged_tuple<type_pair<struct name, std::string>, type_pair<struct age, int>>;
  // it's initialized the same way as a tuple created with the value types of the type pairs (so tuple<string, int> in this case)
  user_t user  { "chris", 21 };
  std::cout << "Name: " << get<name>(user) << std::endl;
  std::cout << "Age: " << get<age>(user) << std::endl;
 // you can still access properties via numeric indexes as if the class was defined as tuple<string, int>
  std::cout << "user[0] = " << get<0>(user) << std::endl;

I opted against having get be a member function to keep it as similar to std::tuple as possible but you could easily add one to the class. Source code here

Here is an implementation similar to ecatmur's answer using the brigand metaprogramming library (https://github.com/edouarda/brigand):

#include <iostream>
#include <brigand/brigand.hpp>

template<typename Members>
class TaggedTuple{

    template<typename Type>
    struct createMember{
        using type = typename Type::second_type;
    };

    using DataTuple = brigand::transform<Members, createMember<brigand::_1>>;
    using Keys = brigand::keys_as_sequence<Members, brigand::list>;
    brigand::as_tuple<DataTuple> members;

public:

    template<typename TagType>
    auto& get(){
        using index = brigand::index_of<Keys, TagType>;
        return std::get<index::value>(members);
    }
};

int main(){

    struct FloatTag{};
    struct IntTag{};
    struct DoubleTag{};

    TaggedTuple<brigand::map<
            brigand::pair<FloatTag, float>,
            brigand::pair<IntTag, int>,
            brigand::pair<DoubleTag, double>>> tagged;

    tagged.get<DoubleTag>() = 200;
    auto val = tagged.get<DoubleTag>();
    std::cout << val << std::endl;

    return 0;
}