How to transform types of boost::fusion::vector?

Question

I need to define two types for specified typelist: first is boost::fusion::vector of these types and second is boost::fusion::vector where references and const are removed for each type in type list.

For example, I have int, unsigned & and long const &. I need to define boost::fusion::vector<int, unsigned &, long const &> and boost::fusion::vector<int, unsigned, long>.

Here is my code:

struct RemoveRef
{
    template <class T>
    struct apply
    {
        using type =
            typename std::remove_const<typename std::remove_reference<T>::type>::type;
    };
};

template <typename...Args>
struct BasicDefinition
{
    typedef typename boost::mpl::vector<Args...> Types;
    typedef typename boost::fusion::result_of::as_vector<Types>::type ArgsType;
    typedef typename boost::mpl::transform<Types, RemoveRef>::type ValueTypes;
    typedef typename boost::fusion::result_of::as_vector<ValueTypes>::type ArgValuesType;
};

It works. I get these types as BasicDefinition<>::ArgsType and BasicDefinition<>::ArgValuesType. But I want to get rid of boost::mpl::vectors and build second type directly from the first one. Is it possible to achieve such result?

Something like:

template <typename...Args>
struct BasicDefinition
{
    typedef typename boost::fusion::vector<Args...> ArgsType;
    typedef ?????<ArgsTypes, RemoveRef>::type ArgValuesType;
};

Answer 1

You might use std::decay_t

template <typename...Args>
struct BasicDefinition
{
    using ArgsType = boost::fusion::vector<Args...>;
    using ArgValuesType = boost::fusion::vector<std::decay_t<Args>...>;
};