CODEWITHSUNDEEP
c++c++11
quant
quant

Reputation: 23052

How do I find out if a tuple contains a type?

Suppose I want to create a compile-time heterogenous container of unique types from some sequence of non-unique types. In order to do this I need to iterate over the source type (some kind of tuple) and check whether each type already exists in my "unique" tuple.

My question is: How can I check whether a tuple (or a boost::fusion container) contains a type?

I'm open to using either the STL or boost.

Upvotes: 43

Views: 15363

Answers (7)

Benno Straub
Benno Straub

Reputation: 2557

C++17 and up solution using fold expressions:

template<typename U, typename... T>
constexpr bool contains(std::tuple<T...>) {
    return (std::is_same_v<U, T> || ...);
}

assert( contains<int   >(std::declval<std::tuple<int, float>>()));
assert( contains<float >(std::declval<std::tuple<int, float>>()));
assert(!contains<double>(std::declval<std::tuple<int, float>>()));

//// In case std::declval ever becomes constexpr, the following should work as well:
// template<typename U, typename Tuple>
// constexpr inline bool contains_v = contains<U>(std::declval<Tuple>());

Upvotes: 12

cdacamara
cdacamara

Reputation: 223

I actually needed something like this for a project. This was my solution:

#include <tuple>
#include <type_traits>

namespace detail {
    struct null { };
}

template <typename T, typename Tuple>
struct tuple_contains;

template <typename T, typename... Ts>
struct tuple_contains<T, std::tuple<Ts...>> :
  std::integral_constant<
    bool,
    !std::is_same<
      std::tuple<typename std::conditional<std::is_same<T, Ts>::value, detail::null, Ts>::type...>,
      std::tuple<Ts...>
    >::value
  >
{ };

The main advantage of this method is that it's one instantiation, no recursion required.

Upvotes: 8

Ross Bencina
Ross Bencina

Reputation: 4173

In C++17 you can do it like this:

template <typename T, typename Tuple>
struct has_type;

template <typename T, typename... Us>
struct has_type<T, std::tuple<Us...>> : std::disjunction<std::is_same<T, Us>...> {};

In C++11 you have to roll your own or / disjunction. Here's a full C++11 version, with tests:

#include <tuple>
#include <type_traits>

template<typename... Conds>
struct or_ : std::false_type {};

template<typename Cond, typename... Conds>
struct or_<Cond, Conds...> : std::conditional<Cond::value, std::true_type, or_<Conds...>>::type
{};

/*
// C++17 version:
template<class... B>
using or_ = std::disjunction<B...>;
*/  

template <typename T, typename Tuple>
struct has_type;

template <typename T, typename... Us>
struct has_type<T, std::tuple<Us...>> : or_<std::is_same<T, Us>...> {};

// Tests
static_assert(has_type<int, std::tuple<>>::value == false, "test");
static_assert(has_type<int, std::tuple<int>>::value == true, "test");
static_assert(has_type<int, std::tuple<float>>::value == false, "test");
static_assert(has_type<int, std::tuple<float, int>>::value == true, "test");
static_assert(has_type<int, std::tuple<int, float>>::value == true, "test");
static_assert(has_type<int, std::tuple<char, float, int>>::value == true, "test");
static_assert(has_type<int, std::tuple<char, float, bool>>::value == false, "test");
static_assert(has_type<const int, std::tuple<int>>::value == false, "test"); // we're using is_same so cv matters
static_assert(has_type<int, std::tuple<const int>>::value == false, "test"); // we're using is_same so cv matters

Upvotes: 28

skypjack
skypjack

Reputation: 50540

Because nobody posted it, I'm adding one more solution based on the bool trick I've learned about here on SO:

#include<type_traits>
#include<tuple>

template<bool...>
struct check {};

template<typename U, typename... T>
constexpr bool contains(std::tuple<T...>) {
    return not std::is_same<
        check<false, std::is_same<U, T>::value...>,
        check<std::is_same<U, T>::value..., false>
    >::value;
}

int main() {
    static_assert(contains<int>(std::tuple<int, char, double>{}), "!");
    static_assert(contains<char>(std::tuple<int, char, double>{}), "!");
    static_assert(contains<double>(std::tuple<int, char, double>{}), "!");
    static_assert(not contains<float>(std::tuple<int, char, double>{}), "!");
    static_assert(not contains<void>(std::tuple<int, char, double>{}), "!");
}

In terms of compile-time performance it's slower than the accepted solution, but it's worth to mention it.

In C++14 it would be even easier to write. The standard template offers already all what you need to do that in the <utility> header:

template<typename U, typename... T>
constexpr auto contains(std::tuple<T...>) {
    return not std::is_same<
        std::integer_sequence<bool, false, std::is_same<U, T>::value...>,
        std::integer_sequence<bool, std::is_same<U, T>::value..., false>
    >::value;
}

This is not far conceptually from what std::get does (available since C++14 for types), but note that the latter fails to compile if the type U is present more than once in T....
If it fits with your requirements mostly depends on the actual problem.

Upvotes: 8

Piotr Skotnicki
Piotr Skotnicki

Reputation: 48447

#include <tuple>
#include <type_traits>

template <typename T, typename Tuple>
struct has_type;

template <typename T>
struct has_type<T, std::tuple<>> : std::false_type {};

template <typename T, typename U, typename... Ts>
struct has_type<T, std::tuple<U, Ts...>> : has_type<T, std::tuple<Ts...>> {};

template <typename T, typename... Ts>
struct has_type<T, std::tuple<T, Ts...>> : std::true_type {};

DEMO

And an additional alias, if the trait itself should be std::true_type or std::false_type :

template <typename T, typename Tuple>
using tuple_contains_type = typename has_type<T, Tuple>::type;

Upvotes: 37

David G
David G

Reputation: 96810

Here is a version that does not recursively instantiate the template to check for a matching type. Instead it uses SFINAE with indices-based meta-programming:

#include <type_traits>
#include <tuple>

template <std::size_t... Indices>
struct index_sequence {
    typedef index_sequence<Indices..., sizeof...(Indices)> next;
};

template <std::size_t Start>
struct make_index_sequence {
    typedef typename make_index_sequence<Start - 1>::type::next type;
};

template <>
struct make_index_sequence<0> {
    typedef index_sequence<> type;
};

template <int n>
using make_index_sequence_t = typename make_index_sequence<n>::type;

template <typename Value, typename Sequence>
struct lookup;

template <typename Value, std::size_t... index>
struct lookup<Value, index_sequence<index...>>
{
private:
    struct null;

    template <typename... Args>
    static std::false_type
    apply(std::conditional_t<std::is_convertible<Args, Value>::value, null, Args>...);

    template <typename...>
    static std::true_type apply(...);

    template <typename... Args>
    static auto apply_helper(Args&&...) ->
    decltype(apply<std::remove_reference_t<Args>...>(std::declval<Args>()...));
public:
    template <typename Tuple>
    using value = decltype(
        apply_helper(
            std::declval<
                typename std::tuple_element<index, Tuple>::type
            >()...
        )
    );
};

template <typename Value, typename Tuple>
using has_type = decltype(
    typename lookup<Value,
                    make_index_sequence_t<std::tuple_size<Tuple>::value>
    >::template value<Tuple>{}
);

Live Demo

Upvotes: 3

pmr
pmr

Reputation: 59811

Since you asked for it, here is a boost::mpl version:

#include <boost/mpl/unique.hpp>
#include <boost/mpl/sort.hpp>
#include <boost/mpl/vector.hpp>
#include <boost/type_traits/is_same.hpp>

using namespace boost;

template<typename Seq>
struct unique_concat : 
  mpl::unique<typename mpl::sort<Seq, is_same<mpl::_1,mpl::_2>>::type, 
              is_same<mpl::_1,mpl::_2>> {};

template<typename T>
struct print;

int main()
{
  typedef mpl::vector<int, float, float, char, int, double, int> input;
  print<unique_concat<input>::type> asdf;

  return 0;
}

Upvotes: 1

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