Can template deduction guides call constexpr functions?












11














I have my own fixed-size array type I want to be constexpr constructible from an std::initializer_list without having to explicitly define the size template argument.



I thought I'd be able to use a template deduction guide but it looks like it's not treating std::initializer_list::size() as a constexpr function for it.



Here's an example of trying to make a deduction guide for std::array (which is similar to my type and has the same problem):



namespace std
{
template<typename T> array(initializer_list<T> initialiserList) -> array<T, initialiserList.size()>;
}
static constexpr std::array myArray = {1,2,3};
static constexpr std::array myArray2 = {{1,2,3}};


I've tried on MSVC and Clang, both give roughly the same errors:
myArray has an error complaining about too many arguments to the function.
myArray2 says "substitution failure [with T = int]: non-type template argument is not a constant expression"



I tried putting constexpr in front of the deduction guide or the function argument but neither appears to be allowed, so it appears that the deduction guide is invalid even though it should work fine in a constexpr context.



Is there a way to make this work without going down the make_array() route?










share|improve this question





























    11














    I have my own fixed-size array type I want to be constexpr constructible from an std::initializer_list without having to explicitly define the size template argument.



    I thought I'd be able to use a template deduction guide but it looks like it's not treating std::initializer_list::size() as a constexpr function for it.



    Here's an example of trying to make a deduction guide for std::array (which is similar to my type and has the same problem):



    namespace std
    {
    template<typename T> array(initializer_list<T> initialiserList) -> array<T, initialiserList.size()>;
    }
    static constexpr std::array myArray = {1,2,3};
    static constexpr std::array myArray2 = {{1,2,3}};


    I've tried on MSVC and Clang, both give roughly the same errors:
    myArray has an error complaining about too many arguments to the function.
    myArray2 says "substitution failure [with T = int]: non-type template argument is not a constant expression"



    I tried putting constexpr in front of the deduction guide or the function argument but neither appears to be allowed, so it appears that the deduction guide is invalid even though it should work fine in a constexpr context.



    Is there a way to make this work without going down the make_array() route?










    share|improve this question



























      11












      11








      11







      I have my own fixed-size array type I want to be constexpr constructible from an std::initializer_list without having to explicitly define the size template argument.



      I thought I'd be able to use a template deduction guide but it looks like it's not treating std::initializer_list::size() as a constexpr function for it.



      Here's an example of trying to make a deduction guide for std::array (which is similar to my type and has the same problem):



      namespace std
      {
      template<typename T> array(initializer_list<T> initialiserList) -> array<T, initialiserList.size()>;
      }
      static constexpr std::array myArray = {1,2,3};
      static constexpr std::array myArray2 = {{1,2,3}};


      I've tried on MSVC and Clang, both give roughly the same errors:
      myArray has an error complaining about too many arguments to the function.
      myArray2 says "substitution failure [with T = int]: non-type template argument is not a constant expression"



      I tried putting constexpr in front of the deduction guide or the function argument but neither appears to be allowed, so it appears that the deduction guide is invalid even though it should work fine in a constexpr context.



      Is there a way to make this work without going down the make_array() route?










      share|improve this question















      I have my own fixed-size array type I want to be constexpr constructible from an std::initializer_list without having to explicitly define the size template argument.



      I thought I'd be able to use a template deduction guide but it looks like it's not treating std::initializer_list::size() as a constexpr function for it.



      Here's an example of trying to make a deduction guide for std::array (which is similar to my type and has the same problem):



      namespace std
      {
      template<typename T> array(initializer_list<T> initialiserList) -> array<T, initialiserList.size()>;
      }
      static constexpr std::array myArray = {1,2,3};
      static constexpr std::array myArray2 = {{1,2,3}};


      I've tried on MSVC and Clang, both give roughly the same errors:
      myArray has an error complaining about too many arguments to the function.
      myArray2 says "substitution failure [with T = int]: non-type template argument is not a constant expression"



      I tried putting constexpr in front of the deduction guide or the function argument but neither appears to be allowed, so it appears that the deduction guide is invalid even though it should work fine in a constexpr context.



      Is there a way to make this work without going down the make_array() route?







      c++ c++17 constexpr template-deduction






      share|improve this question















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      edited Jan 6 at 23:17









      max66

      34.8k63763




      34.8k63763










      asked Jan 6 at 22:53









      Blake PrestonBlake Preston

      835




      835
























          3 Answers
          3






          active

          oldest

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          11














          You can do:



          template <class T, class... U>
          array(T, U...) -> array<T, 1 + sizeof...(U)>;


          The problem is not that you cannot call constexpr functions in deduction guides. You can. This example is ridiculous, but works:



          constexpr size_t plus_one(size_t i) { return i + 1; }

          template <class T, class... U>
          array(T, U...) -> array<T, plus_one(sizeof...(U))>;


          The problem is that function parameters are not constexpr objects, so you cannot invoke constexpr member functions on them if those member functions read kind of local state.






          share|improve this answer





























            8















            Is there a way to make this work without going down the make_array() route?




            Why don't you try with the following deduction guide ?



            template <typename T, std::size_t N>
            array(T const (&)[N]) -> array<T, N>;


            This way, the argument in myArray2 = {{1,2,3}} isn't interpreted as a std::initializer_list (that as argument can't be considered constexpr, so it's size() can't be used for a template argument) but as a C-style array.



            So can be deduced, as template arguments, type and size (T and N) and also the size (N) can be used as template argument.






            share|improve this answer































              2














              Parameter/argument values are not constexpr.



              You might use variadic template to know size at compile time, or type with know size (std::array or C-array reference).






              share|improve this answer





















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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

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                active

                oldest

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                11














                You can do:



                template <class T, class... U>
                array(T, U...) -> array<T, 1 + sizeof...(U)>;


                The problem is not that you cannot call constexpr functions in deduction guides. You can. This example is ridiculous, but works:



                constexpr size_t plus_one(size_t i) { return i + 1; }

                template <class T, class... U>
                array(T, U...) -> array<T, plus_one(sizeof...(U))>;


                The problem is that function parameters are not constexpr objects, so you cannot invoke constexpr member functions on them if those member functions read kind of local state.






                share|improve this answer


























                  11














                  You can do:



                  template <class T, class... U>
                  array(T, U...) -> array<T, 1 + sizeof...(U)>;


                  The problem is not that you cannot call constexpr functions in deduction guides. You can. This example is ridiculous, but works:



                  constexpr size_t plus_one(size_t i) { return i + 1; }

                  template <class T, class... U>
                  array(T, U...) -> array<T, plus_one(sizeof...(U))>;


                  The problem is that function parameters are not constexpr objects, so you cannot invoke constexpr member functions on them if those member functions read kind of local state.






                  share|improve this answer
























                    11












                    11








                    11






                    You can do:



                    template <class T, class... U>
                    array(T, U...) -> array<T, 1 + sizeof...(U)>;


                    The problem is not that you cannot call constexpr functions in deduction guides. You can. This example is ridiculous, but works:



                    constexpr size_t plus_one(size_t i) { return i + 1; }

                    template <class T, class... U>
                    array(T, U...) -> array<T, plus_one(sizeof...(U))>;


                    The problem is that function parameters are not constexpr objects, so you cannot invoke constexpr member functions on them if those member functions read kind of local state.






                    share|improve this answer












                    You can do:



                    template <class T, class... U>
                    array(T, U...) -> array<T, 1 + sizeof...(U)>;


                    The problem is not that you cannot call constexpr functions in deduction guides. You can. This example is ridiculous, but works:



                    constexpr size_t plus_one(size_t i) { return i + 1; }

                    template <class T, class... U>
                    array(T, U...) -> array<T, plus_one(sizeof...(U))>;


                    The problem is that function parameters are not constexpr objects, so you cannot invoke constexpr member functions on them if those member functions read kind of local state.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Jan 6 at 23:22









                    BarryBarry

                    178k18307564




                    178k18307564

























                        8















                        Is there a way to make this work without going down the make_array() route?




                        Why don't you try with the following deduction guide ?



                        template <typename T, std::size_t N>
                        array(T const (&)[N]) -> array<T, N>;


                        This way, the argument in myArray2 = {{1,2,3}} isn't interpreted as a std::initializer_list (that as argument can't be considered constexpr, so it's size() can't be used for a template argument) but as a C-style array.



                        So can be deduced, as template arguments, type and size (T and N) and also the size (N) can be used as template argument.






                        share|improve this answer




























                          8















                          Is there a way to make this work without going down the make_array() route?




                          Why don't you try with the following deduction guide ?



                          template <typename T, std::size_t N>
                          array(T const (&)[N]) -> array<T, N>;


                          This way, the argument in myArray2 = {{1,2,3}} isn't interpreted as a std::initializer_list (that as argument can't be considered constexpr, so it's size() can't be used for a template argument) but as a C-style array.



                          So can be deduced, as template arguments, type and size (T and N) and also the size (N) can be used as template argument.






                          share|improve this answer


























                            8












                            8








                            8







                            Is there a way to make this work without going down the make_array() route?




                            Why don't you try with the following deduction guide ?



                            template <typename T, std::size_t N>
                            array(T const (&)[N]) -> array<T, N>;


                            This way, the argument in myArray2 = {{1,2,3}} isn't interpreted as a std::initializer_list (that as argument can't be considered constexpr, so it's size() can't be used for a template argument) but as a C-style array.



                            So can be deduced, as template arguments, type and size (T and N) and also the size (N) can be used as template argument.






                            share|improve this answer















                            Is there a way to make this work without going down the make_array() route?




                            Why don't you try with the following deduction guide ?



                            template <typename T, std::size_t N>
                            array(T const (&)[N]) -> array<T, N>;


                            This way, the argument in myArray2 = {{1,2,3}} isn't interpreted as a std::initializer_list (that as argument can't be considered constexpr, so it's size() can't be used for a template argument) but as a C-style array.



                            So can be deduced, as template arguments, type and size (T and N) and also the size (N) can be used as template argument.







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Jan 7 at 1:41

























                            answered Jan 6 at 23:10









                            max66max66

                            34.8k63763




                            34.8k63763























                                2














                                Parameter/argument values are not constexpr.



                                You might use variadic template to know size at compile time, or type with know size (std::array or C-array reference).






                                share|improve this answer


























                                  2














                                  Parameter/argument values are not constexpr.



                                  You might use variadic template to know size at compile time, or type with know size (std::array or C-array reference).






                                  share|improve this answer
























                                    2












                                    2








                                    2






                                    Parameter/argument values are not constexpr.



                                    You might use variadic template to know size at compile time, or type with know size (std::array or C-array reference).






                                    share|improve this answer












                                    Parameter/argument values are not constexpr.



                                    You might use variadic template to know size at compile time, or type with know size (std::array or C-array reference).







                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered Jan 6 at 23:07









                                    Jarod42Jarod42

                                    114k12101182




                                    114k12101182






























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