Zero-initializing elements of a std::array with a default member initializer





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11















Suppose I have a class template like this:



template<typename T, size_t N>
struct S {
std::array<T,N> a;
};


Is there a default member initializer I can place on a,



template<typename T, size_t N>
struct S {
std::array<T,N> a = ???;
};


such that no matter what T is, the elements of a will always be initialized (never have indeterminant value)? I.e., even if T is a primitive type like int.










share|improve this question































    11















    Suppose I have a class template like this:



    template<typename T, size_t N>
    struct S {
    std::array<T,N> a;
    };


    Is there a default member initializer I can place on a,



    template<typename T, size_t N>
    struct S {
    std::array<T,N> a = ???;
    };


    such that no matter what T is, the elements of a will always be initialized (never have indeterminant value)? I.e., even if T is a primitive type like int.










    share|improve this question



























      11












      11








      11


      1






      Suppose I have a class template like this:



      template<typename T, size_t N>
      struct S {
      std::array<T,N> a;
      };


      Is there a default member initializer I can place on a,



      template<typename T, size_t N>
      struct S {
      std::array<T,N> a = ???;
      };


      such that no matter what T is, the elements of a will always be initialized (never have indeterminant value)? I.e., even if T is a primitive type like int.










      share|improve this question
















      Suppose I have a class template like this:



      template<typename T, size_t N>
      struct S {
      std::array<T,N> a;
      };


      Is there a default member initializer I can place on a,



      template<typename T, size_t N>
      struct S {
      std::array<T,N> a = ???;
      };


      such that no matter what T is, the elements of a will always be initialized (never have indeterminant value)? I.e., even if T is a primitive type like int.







      c++ c++17






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Jan 6 at 0:25









      Peter Mortensen

      13.9k1987113




      13.9k1987113










      asked Jan 5 at 13:55









      Andrew TomazosAndrew Tomazos

      35.7k26134234




      35.7k26134234
























          2 Answers
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          18














          This:



          template<typename T, size_t N>
          struct S {
          std::array<T,N> a = {};
          };


          That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:



          struct A {
          A() {}
          int i;
          };


          which would prevent i from being initialized. But that's on them.






          share|improve this answer

































            6














            std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.






            share|improve this answer


























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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

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              active

              oldest

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              18














              This:



              template<typename T, size_t N>
              struct S {
              std::array<T,N> a = {};
              };


              That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:



              struct A {
              A() {}
              int i;
              };


              which would prevent i from being initialized. But that's on them.






              share|improve this answer






























                18














                This:



                template<typename T, size_t N>
                struct S {
                std::array<T,N> a = {};
                };


                That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:



                struct A {
                A() {}
                int i;
                };


                which would prevent i from being initialized. But that's on them.






                share|improve this answer




























                  18












                  18








                  18







                  This:



                  template<typename T, size_t N>
                  struct S {
                  std::array<T,N> a = {};
                  };


                  That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:



                  struct A {
                  A() {}
                  int i;
                  };


                  which would prevent i from being initialized. But that's on them.






                  share|improve this answer















                  This:



                  template<typename T, size_t N>
                  struct S {
                  std::array<T,N> a = {};
                  };


                  That will recursively copy-initialize each element from {}. For int, that will zero-initialize. Of course, someone can always write:



                  struct A {
                  A() {}
                  int i;
                  };


                  which would prevent i from being initialized. But that's on them.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Jan 5 at 14:10

























                  answered Jan 5 at 13:59









                  BarryBarry

                  186k21329606




                  186k21329606

























                      6














                      std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.






                      share|improve this answer






























                        6














                        std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.






                        share|improve this answer




























                          6












                          6








                          6







                          std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.






                          share|improve this answer















                          std::array is an aggregate type. You can aggregate initialize it with empty braces {} and that will initialize accordingly the elements of the internal array of T that std::array holds.







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Jan 5 at 14:25

























                          answered Jan 5 at 14:00









                          JansJans

                          9,19422735




                          9,19422735






























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