CODEWITHSUNDEEP
c++templatestemplate-specializationtype-traits
Cory
Cory

Reputation: 161

Find template type of a template type c++

I would like to have a function that can take many different things (for simplicity) like so:

template <typename T>
typename type_to_return<T>::type   // <-- Use type_to_return to get result type
foo(T t)
{
    return typename type_to_return<T>::type(T); // <-- Build a thing!
}

I would then specialize the type_to_return class for the types I have created. This would make the entry be one function and I could then just define new type_to_returns and constructors.

I want type_to_return<T>::type to be just T if T is not some class template. Otherwise I want it to be that class's first template parameter. So for int, I get back int, and for MultOp<float,int,double>, I want to get back float.

How do I do that? I think I need to do something like:

// Base version
template <typename T>
struct type_to_return
{
    typedef T type;
};

// Specialized type
template <template <typename> class T>
struct type_to_return<T <any_type_somehow> >
{
    typedef template boost::magic_type_unwrapper<T>::param<1>::type type;
};

Upvotes: 1

Views: 1281

Answers (1)

Jarod42
Jarod42

Reputation: 217075

You may implement a type_unwrapper as follow:

template <typename T>
struct type_unwrapper;

template <template <typename...> class C, typename... Ts>
struct type_unwrapper<C<Ts...>>
{
    static constexpr std::size_t type_count = sizeof...(Ts);

    template <std::size_t N>
    using param_t = typename std::tuple_element<N, std::tuple<Ts...>>::type;
};

which works as long there is no template value as in std::array<T, N>.

Note also that stl container declare some typedef to retrieve there template arguments as std::vector<T, Alloc>::value_type which is T.

Upvotes: 4

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