How to use the using keyword for a variation pattern

I have a variadic templated class that can accept any number of variables in a constructor, std::tuple/std::pair

and so on.

I would like to use this wrapper for functions with different return types.

For example:

class f1
{
    using output = double;
    output operator() { do_smth };
}
class f2
{
    using output = std::tuple<double,int>;
    output operator() { do_smth };
}



template <typename... Types>
class OutputType
{
    std::tuple<Types...> m_val;
public:
    OutputType(std::tuple<Types...>&& val) : m_val(val) {};
    OutputType(Types&& ... val) : m_val(std::forward<Types>(Types)...) {};
};

      

Now in the third class, I would like to declare the following:

template <typename F>
class dummy
{
    using Output = typename OutputType(typename F::Output));
}

      

How do I declare the above statement so that it does the right thing for and for dummy<f2>

?

(i.e. OutputType<double,int>

, not OutputType<std::tuple<double,int>>

)

+3


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2 answers


If I understand your question correctly (?), You can define type traits as

template <typename ... Types>
struct oTypes
 { using type = OutputType<Types...>; };

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

      

and then define dummy

like this



template <typename F>
struct dummy
 { using output = typename oTypes<typename F::output>::type; };

      

Below is a complete compiled example

#include <tuple>
#include <utility>

struct f1
 {
   using output = double;

   output operator() ()
    { return 0.0; }
 };

struct f2
 {
   using output = std::tuple<double,int>;

   output operator() ()
    { return { 1.0, 2 }; }
 };

template <typename ... Types>
class OutputType
 {
   private:
      std::tuple<Types...> m_val;

   public:
      OutputType(std::tuple<Types...>&& val) : m_val(val)
       { }

      OutputType(Types&& ... val) : m_val(std::forward<Types>(val)...)
       { }
 };

template <typename ... Types>
struct oTypes
 { using type = OutputType<Types...>; };

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

template <typename F>
struct dummy
 { using output = typename oTypes<typename F::output>::type; };

int main()
 { 
   static_assert( std::is_same<dummy<f1>::output,
                               OutputType<double>>::value, "!");
   static_assert( std::is_same<dummy<f2>::output,
                               OutputType<double, int>>::value, "!!");
 }

      

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A helper template like

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

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

      



Which changes the OutputType to something like

template <typename... Types>
class OutputType
{
    typename add_tuple<Types...>::type m_val;
public:
    OutputType(typename add_tuple<Types...>::type&& val) : m_val(val) {};
    OutputType(Types&& ... val) : m_val(std::forward<Types>(Types)...) {};
};

      

0


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