Checking if variadic template parameters are unique using fold expressions












6














Given a variadic template parameter pack, I want to check if all types given to it are unique using an inline constexpr bool and fold expressions. I trie something like this:



template<class... T>
inline static constexpr bool is_unique = (... && (!is_one_of<T, ...>));


Where is_one_of is a similar bool that works correctly.
But this line doesn't compile regardless of what I put into is_one_of. Can this even be done using fold expressions, or do I need to use a regular struct for this purpose?










share|improve this question





























    6














    Given a variadic template parameter pack, I want to check if all types given to it are unique using an inline constexpr bool and fold expressions. I trie something like this:



    template<class... T>
    inline static constexpr bool is_unique = (... && (!is_one_of<T, ...>));


    Where is_one_of is a similar bool that works correctly.
    But this line doesn't compile regardless of what I put into is_one_of. Can this even be done using fold expressions, or do I need to use a regular struct for this purpose?










    share|improve this question



























      6












      6








      6


      2





      Given a variadic template parameter pack, I want to check if all types given to it are unique using an inline constexpr bool and fold expressions. I trie something like this:



      template<class... T>
      inline static constexpr bool is_unique = (... && (!is_one_of<T, ...>));


      Where is_one_of is a similar bool that works correctly.
      But this line doesn't compile regardless of what I put into is_one_of. Can this even be done using fold expressions, or do I need to use a regular struct for this purpose?










      share|improve this question















      Given a variadic template parameter pack, I want to check if all types given to it are unique using an inline constexpr bool and fold expressions. I trie something like this:



      template<class... T>
      inline static constexpr bool is_unique = (... && (!is_one_of<T, ...>));


      Where is_one_of is a similar bool that works correctly.
      But this line doesn't compile regardless of what I put into is_one_of. Can this even be done using fold expressions, or do I need to use a regular struct for this purpose?







      c++ templates c++17 variadic-templates fold-expression






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 21 '18 at 13:56









      max66

      34.5k63762




      34.5k63762










      asked Nov 27 '17 at 12:50









      Joald

      391216




      391216
























          2 Answers
          2






          active

          oldest

          votes


















          9














          You approach doesn't really work because is_one_of needs to be called with a type T and all the remaining types not including T. There's no way of expressing that with a fold expression over a single parameter pack. I suggest using specialization instead:



          template <typename...>
          inline constexpr auto is_unique = std::true_type{};

          template <typename T, typename... Rest>
          inline constexpr auto is_unique<T, Rest...> = std::bool_constant<
          (!std::is_same_v<T, Rest> && ...) && is_unique<Rest...>
          >{};


          Usage:



          static_assert(is_unique<>);
          static_assert(is_unique<int>);
          static_assert(is_unique<int, float, double>);
          static_assert(!is_unique<int, float, double, int>);


          live example on wandbox.org





          (Thanks to Barry for the simplification that uses a fold expression.)






          share|improve this answer























          • You're right. I trusted my YouCompleteMe config too much, which surprisingly gave me diags even though it didn't even grok all the c++17 features. Disregard. I'll have to wrap my head around why it works, though because I suspected the given case to fail, from brain-compiling it. Got it; My brain-compiler failed to see T1, Rest... as "just the tail", which it of course is. +1
            – sehe
            Nov 27 '17 at 13:05








          • 3




            Could simplify to T0, Rest... with the body of (!is_same_v<T0, Rest> && ...) && is_unique<Rest...>?
            – Barry
            Nov 27 '17 at 16:27






          • 2




            While the other answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so your answer "You can't do it" is more appropriate. Thanks!
            – Joald
            Nov 27 '17 at 19:21





















          1














          -- EDIT --



          googling I've found an interesting solution that give me inspiration to avoid recursion and to avoid a lot of warnings



          So you can define a wrapper of type



          template <typename>
          struct wrapT
          { };


          and a wrapper for type and integer that inherit from the wrapper for type



          template <typename T, std::size_t>
          struct wrapTI : public wrapT<T>
          { };


          Next you can define a foo class that recursively inherit from wrapTI



          template <typename T,
          typename = std::make_index_sequence<std::tuple_size<T>::value>>
          struct foo;

          template <typename ... Ts, std::size_t ... Is>
          struct foo<std::tuple<Ts...>, std::index_sequence<Is...>>
          : public wrapTI<Ts, Is>...
          { };


          Now is_unique can be something like



          template <typename ... Ts>
          static constexpr bool isUnique
          = ( ... && std::is_convertible<foo<std::tuple<Ts...>>, wrapT<Ts>>::value );


          The point is that foo<Ts...> can be converted to wrapT<T> only if foo<Ts...> inherit one time (and only one time) from wrapT<T>, that is if T is present one time (and only one time) in Ts....



          The following is a full compiling example



          #include <tuple>
          #include <type_traits>

          template <typename>
          struct wrapT
          { };

          template <typename T, std::size_t>
          struct wrapTI : public wrapT<T>
          { };

          template <typename T,
          typename = std::make_index_sequence<std::tuple_size<T>::value>>
          struct foo;

          template <typename ... Ts, std::size_t ... Is>
          struct foo<std::tuple<Ts...>, std::index_sequence<Is...>>
          : public wrapTI<Ts, Is>...
          { };

          template <typename ... Ts>
          static constexpr bool isUnique
          = ( ... && std::is_convertible<foo<std::tuple<Ts...>>, wrapT<Ts>>::value );

          int main ()
          {
          static_assert( true == isUnique<int, long, long long> );
          static_assert( false == isUnique<int, long, long long, int> );
          }





          share|improve this answer























          • While your answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so the other answer "You can't do it" is more appropriate. But thanks, cool to know that it can be done!
            – Joald
            Nov 27 '17 at 19:22










          • @Joald - I'm agree with you: the Vittorio's answer is better than mine (and also I find annoying those warnings)
            – max66
            Nov 27 '17 at 21:08










          • @Joald - Just for fun... answer modified, recursion avoided and warnings removed :) - But added another wrapper struct :(
            – max66
            Nov 27 '17 at 22:12











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          2 Answers
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          active

          oldest

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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          9














          You approach doesn't really work because is_one_of needs to be called with a type T and all the remaining types not including T. There's no way of expressing that with a fold expression over a single parameter pack. I suggest using specialization instead:



          template <typename...>
          inline constexpr auto is_unique = std::true_type{};

          template <typename T, typename... Rest>
          inline constexpr auto is_unique<T, Rest...> = std::bool_constant<
          (!std::is_same_v<T, Rest> && ...) && is_unique<Rest...>
          >{};


          Usage:



          static_assert(is_unique<>);
          static_assert(is_unique<int>);
          static_assert(is_unique<int, float, double>);
          static_assert(!is_unique<int, float, double, int>);


          live example on wandbox.org





          (Thanks to Barry for the simplification that uses a fold expression.)






          share|improve this answer























          • You're right. I trusted my YouCompleteMe config too much, which surprisingly gave me diags even though it didn't even grok all the c++17 features. Disregard. I'll have to wrap my head around why it works, though because I suspected the given case to fail, from brain-compiling it. Got it; My brain-compiler failed to see T1, Rest... as "just the tail", which it of course is. +1
            – sehe
            Nov 27 '17 at 13:05








          • 3




            Could simplify to T0, Rest... with the body of (!is_same_v<T0, Rest> && ...) && is_unique<Rest...>?
            – Barry
            Nov 27 '17 at 16:27






          • 2




            While the other answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so your answer "You can't do it" is more appropriate. Thanks!
            – Joald
            Nov 27 '17 at 19:21


















          9














          You approach doesn't really work because is_one_of needs to be called with a type T and all the remaining types not including T. There's no way of expressing that with a fold expression over a single parameter pack. I suggest using specialization instead:



          template <typename...>
          inline constexpr auto is_unique = std::true_type{};

          template <typename T, typename... Rest>
          inline constexpr auto is_unique<T, Rest...> = std::bool_constant<
          (!std::is_same_v<T, Rest> && ...) && is_unique<Rest...>
          >{};


          Usage:



          static_assert(is_unique<>);
          static_assert(is_unique<int>);
          static_assert(is_unique<int, float, double>);
          static_assert(!is_unique<int, float, double, int>);


          live example on wandbox.org





          (Thanks to Barry for the simplification that uses a fold expression.)






          share|improve this answer























          • You're right. I trusted my YouCompleteMe config too much, which surprisingly gave me diags even though it didn't even grok all the c++17 features. Disregard. I'll have to wrap my head around why it works, though because I suspected the given case to fail, from brain-compiling it. Got it; My brain-compiler failed to see T1, Rest... as "just the tail", which it of course is. +1
            – sehe
            Nov 27 '17 at 13:05








          • 3




            Could simplify to T0, Rest... with the body of (!is_same_v<T0, Rest> && ...) && is_unique<Rest...>?
            – Barry
            Nov 27 '17 at 16:27






          • 2




            While the other answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so your answer "You can't do it" is more appropriate. Thanks!
            – Joald
            Nov 27 '17 at 19:21
















          9












          9








          9






          You approach doesn't really work because is_one_of needs to be called with a type T and all the remaining types not including T. There's no way of expressing that with a fold expression over a single parameter pack. I suggest using specialization instead:



          template <typename...>
          inline constexpr auto is_unique = std::true_type{};

          template <typename T, typename... Rest>
          inline constexpr auto is_unique<T, Rest...> = std::bool_constant<
          (!std::is_same_v<T, Rest> && ...) && is_unique<Rest...>
          >{};


          Usage:



          static_assert(is_unique<>);
          static_assert(is_unique<int>);
          static_assert(is_unique<int, float, double>);
          static_assert(!is_unique<int, float, double, int>);


          live example on wandbox.org





          (Thanks to Barry for the simplification that uses a fold expression.)






          share|improve this answer














          You approach doesn't really work because is_one_of needs to be called with a type T and all the remaining types not including T. There's no way of expressing that with a fold expression over a single parameter pack. I suggest using specialization instead:



          template <typename...>
          inline constexpr auto is_unique = std::true_type{};

          template <typename T, typename... Rest>
          inline constexpr auto is_unique<T, Rest...> = std::bool_constant<
          (!std::is_same_v<T, Rest> && ...) && is_unique<Rest...>
          >{};


          Usage:



          static_assert(is_unique<>);
          static_assert(is_unique<int>);
          static_assert(is_unique<int, float, double>);
          static_assert(!is_unique<int, float, double, int>);


          live example on wandbox.org





          (Thanks to Barry for the simplification that uses a fold expression.)







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 27 '17 at 22:51

























          answered Nov 27 '17 at 12:56









          Vittorio Romeo

          57.4k17154293




          57.4k17154293












          • You're right. I trusted my YouCompleteMe config too much, which surprisingly gave me diags even though it didn't even grok all the c++17 features. Disregard. I'll have to wrap my head around why it works, though because I suspected the given case to fail, from brain-compiling it. Got it; My brain-compiler failed to see T1, Rest... as "just the tail", which it of course is. +1
            – sehe
            Nov 27 '17 at 13:05








          • 3




            Could simplify to T0, Rest... with the body of (!is_same_v<T0, Rest> && ...) && is_unique<Rest...>?
            – Barry
            Nov 27 '17 at 16:27






          • 2




            While the other answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so your answer "You can't do it" is more appropriate. Thanks!
            – Joald
            Nov 27 '17 at 19:21




















          • You're right. I trusted my YouCompleteMe config too much, which surprisingly gave me diags even though it didn't even grok all the c++17 features. Disregard. I'll have to wrap my head around why it works, though because I suspected the given case to fail, from brain-compiling it. Got it; My brain-compiler failed to see T1, Rest... as "just the tail", which it of course is. +1
            – sehe
            Nov 27 '17 at 13:05








          • 3




            Could simplify to T0, Rest... with the body of (!is_same_v<T0, Rest> && ...) && is_unique<Rest...>?
            – Barry
            Nov 27 '17 at 16:27






          • 2




            While the other answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so your answer "You can't do it" is more appropriate. Thanks!
            – Joald
            Nov 27 '17 at 19:21


















          You're right. I trusted my YouCompleteMe config too much, which surprisingly gave me diags even though it didn't even grok all the c++17 features. Disregard. I'll have to wrap my head around why it works, though because I suspected the given case to fail, from brain-compiling it. Got it; My brain-compiler failed to see T1, Rest... as "just the tail", which it of course is. +1
          – sehe
          Nov 27 '17 at 13:05






          You're right. I trusted my YouCompleteMe config too much, which surprisingly gave me diags even though it didn't even grok all the c++17 features. Disregard. I'll have to wrap my head around why it works, though because I suspected the given case to fail, from brain-compiling it. Got it; My brain-compiler failed to see T1, Rest... as "just the tail", which it of course is. +1
          – sehe
          Nov 27 '17 at 13:05






          3




          3




          Could simplify to T0, Rest... with the body of (!is_same_v<T0, Rest> && ...) && is_unique<Rest...>?
          – Barry
          Nov 27 '17 at 16:27




          Could simplify to T0, Rest... with the body of (!is_same_v<T0, Rest> && ...) && is_unique<Rest...>?
          – Barry
          Nov 27 '17 at 16:27




          2




          2




          While the other answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so your answer "You can't do it" is more appropriate. Thanks!
          – Joald
          Nov 27 '17 at 19:21






          While the other answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so your answer "You can't do it" is more appropriate. Thanks!
          – Joald
          Nov 27 '17 at 19:21















          1














          -- EDIT --



          googling I've found an interesting solution that give me inspiration to avoid recursion and to avoid a lot of warnings



          So you can define a wrapper of type



          template <typename>
          struct wrapT
          { };


          and a wrapper for type and integer that inherit from the wrapper for type



          template <typename T, std::size_t>
          struct wrapTI : public wrapT<T>
          { };


          Next you can define a foo class that recursively inherit from wrapTI



          template <typename T,
          typename = std::make_index_sequence<std::tuple_size<T>::value>>
          struct foo;

          template <typename ... Ts, std::size_t ... Is>
          struct foo<std::tuple<Ts...>, std::index_sequence<Is...>>
          : public wrapTI<Ts, Is>...
          { };


          Now is_unique can be something like



          template <typename ... Ts>
          static constexpr bool isUnique
          = ( ... && std::is_convertible<foo<std::tuple<Ts...>>, wrapT<Ts>>::value );


          The point is that foo<Ts...> can be converted to wrapT<T> only if foo<Ts...> inherit one time (and only one time) from wrapT<T>, that is if T is present one time (and only one time) in Ts....



          The following is a full compiling example



          #include <tuple>
          #include <type_traits>

          template <typename>
          struct wrapT
          { };

          template <typename T, std::size_t>
          struct wrapTI : public wrapT<T>
          { };

          template <typename T,
          typename = std::make_index_sequence<std::tuple_size<T>::value>>
          struct foo;

          template <typename ... Ts, std::size_t ... Is>
          struct foo<std::tuple<Ts...>, std::index_sequence<Is...>>
          : public wrapTI<Ts, Is>...
          { };

          template <typename ... Ts>
          static constexpr bool isUnique
          = ( ... && std::is_convertible<foo<std::tuple<Ts...>>, wrapT<Ts>>::value );

          int main ()
          {
          static_assert( true == isUnique<int, long, long long> );
          static_assert( false == isUnique<int, long, long long, int> );
          }





          share|improve this answer























          • While your answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so the other answer "You can't do it" is more appropriate. But thanks, cool to know that it can be done!
            – Joald
            Nov 27 '17 at 19:22










          • @Joald - I'm agree with you: the Vittorio's answer is better than mine (and also I find annoying those warnings)
            – max66
            Nov 27 '17 at 21:08










          • @Joald - Just for fun... answer modified, recursion avoided and warnings removed :) - But added another wrapper struct :(
            – max66
            Nov 27 '17 at 22:12
















          1














          -- EDIT --



          googling I've found an interesting solution that give me inspiration to avoid recursion and to avoid a lot of warnings



          So you can define a wrapper of type



          template <typename>
          struct wrapT
          { };


          and a wrapper for type and integer that inherit from the wrapper for type



          template <typename T, std::size_t>
          struct wrapTI : public wrapT<T>
          { };


          Next you can define a foo class that recursively inherit from wrapTI



          template <typename T,
          typename = std::make_index_sequence<std::tuple_size<T>::value>>
          struct foo;

          template <typename ... Ts, std::size_t ... Is>
          struct foo<std::tuple<Ts...>, std::index_sequence<Is...>>
          : public wrapTI<Ts, Is>...
          { };


          Now is_unique can be something like



          template <typename ... Ts>
          static constexpr bool isUnique
          = ( ... && std::is_convertible<foo<std::tuple<Ts...>>, wrapT<Ts>>::value );


          The point is that foo<Ts...> can be converted to wrapT<T> only if foo<Ts...> inherit one time (and only one time) from wrapT<T>, that is if T is present one time (and only one time) in Ts....



          The following is a full compiling example



          #include <tuple>
          #include <type_traits>

          template <typename>
          struct wrapT
          { };

          template <typename T, std::size_t>
          struct wrapTI : public wrapT<T>
          { };

          template <typename T,
          typename = std::make_index_sequence<std::tuple_size<T>::value>>
          struct foo;

          template <typename ... Ts, std::size_t ... Is>
          struct foo<std::tuple<Ts...>, std::index_sequence<Is...>>
          : public wrapTI<Ts, Is>...
          { };

          template <typename ... Ts>
          static constexpr bool isUnique
          = ( ... && std::is_convertible<foo<std::tuple<Ts...>>, wrapT<Ts>>::value );

          int main ()
          {
          static_assert( true == isUnique<int, long, long long> );
          static_assert( false == isUnique<int, long, long long, int> );
          }





          share|improve this answer























          • While your answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so the other answer "You can't do it" is more appropriate. But thanks, cool to know that it can be done!
            – Joald
            Nov 27 '17 at 19:22










          • @Joald - I'm agree with you: the Vittorio's answer is better than mine (and also I find annoying those warnings)
            – max66
            Nov 27 '17 at 21:08










          • @Joald - Just for fun... answer modified, recursion avoided and warnings removed :) - But added another wrapper struct :(
            – max66
            Nov 27 '17 at 22:12














          1












          1








          1






          -- EDIT --



          googling I've found an interesting solution that give me inspiration to avoid recursion and to avoid a lot of warnings



          So you can define a wrapper of type



          template <typename>
          struct wrapT
          { };


          and a wrapper for type and integer that inherit from the wrapper for type



          template <typename T, std::size_t>
          struct wrapTI : public wrapT<T>
          { };


          Next you can define a foo class that recursively inherit from wrapTI



          template <typename T,
          typename = std::make_index_sequence<std::tuple_size<T>::value>>
          struct foo;

          template <typename ... Ts, std::size_t ... Is>
          struct foo<std::tuple<Ts...>, std::index_sequence<Is...>>
          : public wrapTI<Ts, Is>...
          { };


          Now is_unique can be something like



          template <typename ... Ts>
          static constexpr bool isUnique
          = ( ... && std::is_convertible<foo<std::tuple<Ts...>>, wrapT<Ts>>::value );


          The point is that foo<Ts...> can be converted to wrapT<T> only if foo<Ts...> inherit one time (and only one time) from wrapT<T>, that is if T is present one time (and only one time) in Ts....



          The following is a full compiling example



          #include <tuple>
          #include <type_traits>

          template <typename>
          struct wrapT
          { };

          template <typename T, std::size_t>
          struct wrapTI : public wrapT<T>
          { };

          template <typename T,
          typename = std::make_index_sequence<std::tuple_size<T>::value>>
          struct foo;

          template <typename ... Ts, std::size_t ... Is>
          struct foo<std::tuple<Ts...>, std::index_sequence<Is...>>
          : public wrapTI<Ts, Is>...
          { };

          template <typename ... Ts>
          static constexpr bool isUnique
          = ( ... && std::is_convertible<foo<std::tuple<Ts...>>, wrapT<Ts>>::value );

          int main ()
          {
          static_assert( true == isUnique<int, long, long long> );
          static_assert( false == isUnique<int, long, long long, int> );
          }





          share|improve this answer














          -- EDIT --



          googling I've found an interesting solution that give me inspiration to avoid recursion and to avoid a lot of warnings



          So you can define a wrapper of type



          template <typename>
          struct wrapT
          { };


          and a wrapper for type and integer that inherit from the wrapper for type



          template <typename T, std::size_t>
          struct wrapTI : public wrapT<T>
          { };


          Next you can define a foo class that recursively inherit from wrapTI



          template <typename T,
          typename = std::make_index_sequence<std::tuple_size<T>::value>>
          struct foo;

          template <typename ... Ts, std::size_t ... Is>
          struct foo<std::tuple<Ts...>, std::index_sequence<Is...>>
          : public wrapTI<Ts, Is>...
          { };


          Now is_unique can be something like



          template <typename ... Ts>
          static constexpr bool isUnique
          = ( ... && std::is_convertible<foo<std::tuple<Ts...>>, wrapT<Ts>>::value );


          The point is that foo<Ts...> can be converted to wrapT<T> only if foo<Ts...> inherit one time (and only one time) from wrapT<T>, that is if T is present one time (and only one time) in Ts....



          The following is a full compiling example



          #include <tuple>
          #include <type_traits>

          template <typename>
          struct wrapT
          { };

          template <typename T, std::size_t>
          struct wrapTI : public wrapT<T>
          { };

          template <typename T,
          typename = std::make_index_sequence<std::tuple_size<T>::value>>
          struct foo;

          template <typename ... Ts, std::size_t ... Is>
          struct foo<std::tuple<Ts...>, std::index_sequence<Is...>>
          : public wrapTI<Ts, Is>...
          { };

          template <typename ... Ts>
          static constexpr bool isUnique
          = ( ... && std::is_convertible<foo<std::tuple<Ts...>>, wrapT<Ts>>::value );

          int main ()
          {
          static_assert( true == isUnique<int, long, long long> );
          static_assert( false == isUnique<int, long, long long, int> );
          }






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 27 '17 at 22:10

























          answered Nov 27 '17 at 14:23









          max66

          34.5k63762




          34.5k63762












          • While your answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so the other answer "You can't do it" is more appropriate. But thanks, cool to know that it can be done!
            – Joald
            Nov 27 '17 at 19:22










          • @Joald - I'm agree with you: the Vittorio's answer is better than mine (and also I find annoying those warnings)
            – max66
            Nov 27 '17 at 21:08










          • @Joald - Just for fun... answer modified, recursion avoided and warnings removed :) - But added another wrapper struct :(
            – max66
            Nov 27 '17 at 22:12


















          • While your answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so the other answer "You can't do it" is more appropriate. But thanks, cool to know that it can be done!
            – Joald
            Nov 27 '17 at 19:22










          • @Joald - I'm agree with you: the Vittorio's answer is better than mine (and also I find annoying those warnings)
            – max66
            Nov 27 '17 at 21:08










          • @Joald - Just for fun... answer modified, recursion avoided and warnings removed :) - But added another wrapper struct :(
            – max66
            Nov 27 '17 at 22:12
















          While your answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so the other answer "You can't do it" is more appropriate. But thanks, cool to know that it can be done!
          – Joald
          Nov 27 '17 at 19:22




          While your answer provides a way to do that with fold expression and constexpr bool as I wanted, it comes with a lot of warnings and still requires additional structs which I wanted to do without, so the other answer "You can't do it" is more appropriate. But thanks, cool to know that it can be done!
          – Joald
          Nov 27 '17 at 19:22












          @Joald - I'm agree with you: the Vittorio's answer is better than mine (and also I find annoying those warnings)
          – max66
          Nov 27 '17 at 21:08




          @Joald - I'm agree with you: the Vittorio's answer is better than mine (and also I find annoying those warnings)
          – max66
          Nov 27 '17 at 21:08












          @Joald - Just for fun... answer modified, recursion avoided and warnings removed :) - But added another wrapper struct :(
          – max66
          Nov 27 '17 at 22:12




          @Joald - Just for fun... answer modified, recursion avoided and warnings removed :) - But added another wrapper struct :(
          – max66
          Nov 27 '17 at 22:12


















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