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Template specialization on array pointer with any layer of nesting



Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manara
Data science time! April 2019 and salary with experience
The Ask Question Wizard is Live!Template specialization to use default type if class member typedef does not existPretty-print C++ STL containersReference level part 2Query about C++ template specialization and partial template specializationUsing const char** with Template SpecializationFunction template specialization - problems with pointersReplacing a 32-bit loop counter with 64-bit introduces crazy performance deviationspartial template specialization for template pointer to functionconstexpr array member with template specialization: inconsistent behavior cross compilersDelete templated struct partial specialization



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1















I need to write a template specialization which will work with pointers on arrays (these things: char(*)[]). And we will write this code for that



// our class
template<typename T>
struct CoolStruct

static void Print(); // will print "Default"
;

//one specialization for sized arrays
template<typename T, size_t S>
struct CoolStruct<T(*)[S]>

static void Print(); // will print "T(*)[sized]"
;

// and one for arrays without size
template<typename T>
struct CoolStruct<T(*)[]>

static void Print(); // will print "T(*)[]"
;


And when in our code we will do something like this:



int (*arrptr)[10];
CoolClass<decltype(arrptr)>::Print();


The console will print "T(*)[sized]" as we expected (considering that we wrote implementation for all methods of course). But what if we will wrote our code like this:



int (**arrptr_d)[10];
CoolClass<decltype(arrptr_d)>::Print();


In this case the console will actually print "Default". We could write another specialization like this:



template<typename T, size_t S>
struct CoolStruct<T(**)[S]>

static void Print(); // will print "T(*)[sized]"
;


but I want to find another solution (I don't want to write another dozen specializations when I would need to use int(****)[]). So considering we using C++17 standard, is there a way to do such a thing?



P.S. And yes, sorry for my terrible English; it is not my native language.










share|improve this question






























    1















    I need to write a template specialization which will work with pointers on arrays (these things: char(*)[]). And we will write this code for that



    // our class
    template<typename T>
    struct CoolStruct

    static void Print(); // will print "Default"
    ;

    //one specialization for sized arrays
    template<typename T, size_t S>
    struct CoolStruct<T(*)[S]>

    static void Print(); // will print "T(*)[sized]"
    ;

    // and one for arrays without size
    template<typename T>
    struct CoolStruct<T(*)[]>

    static void Print(); // will print "T(*)[]"
    ;


    And when in our code we will do something like this:



    int (*arrptr)[10];
    CoolClass<decltype(arrptr)>::Print();


    The console will print "T(*)[sized]" as we expected (considering that we wrote implementation for all methods of course). But what if we will wrote our code like this:



    int (**arrptr_d)[10];
    CoolClass<decltype(arrptr_d)>::Print();


    In this case the console will actually print "Default". We could write another specialization like this:



    template<typename T, size_t S>
    struct CoolStruct<T(**)[S]>

    static void Print(); // will print "T(*)[sized]"
    ;


    but I want to find another solution (I don't want to write another dozen specializations when I would need to use int(****)[]). So considering we using C++17 standard, is there a way to do such a thing?



    P.S. And yes, sorry for my terrible English; it is not my native language.










    share|improve this question


























      1












      1








      1








      I need to write a template specialization which will work with pointers on arrays (these things: char(*)[]). And we will write this code for that



      // our class
      template<typename T>
      struct CoolStruct

      static void Print(); // will print "Default"
      ;

      //one specialization for sized arrays
      template<typename T, size_t S>
      struct CoolStruct<T(*)[S]>

      static void Print(); // will print "T(*)[sized]"
      ;

      // and one for arrays without size
      template<typename T>
      struct CoolStruct<T(*)[]>

      static void Print(); // will print "T(*)[]"
      ;


      And when in our code we will do something like this:



      int (*arrptr)[10];
      CoolClass<decltype(arrptr)>::Print();


      The console will print "T(*)[sized]" as we expected (considering that we wrote implementation for all methods of course). But what if we will wrote our code like this:



      int (**arrptr_d)[10];
      CoolClass<decltype(arrptr_d)>::Print();


      In this case the console will actually print "Default". We could write another specialization like this:



      template<typename T, size_t S>
      struct CoolStruct<T(**)[S]>

      static void Print(); // will print "T(*)[sized]"
      ;


      but I want to find another solution (I don't want to write another dozen specializations when I would need to use int(****)[]). So considering we using C++17 standard, is there a way to do such a thing?



      P.S. And yes, sorry for my terrible English; it is not my native language.










      share|improve this question
















      I need to write a template specialization which will work with pointers on arrays (these things: char(*)[]). And we will write this code for that



      // our class
      template<typename T>
      struct CoolStruct

      static void Print(); // will print "Default"
      ;

      //one specialization for sized arrays
      template<typename T, size_t S>
      struct CoolStruct<T(*)[S]>

      static void Print(); // will print "T(*)[sized]"
      ;

      // and one for arrays without size
      template<typename T>
      struct CoolStruct<T(*)[]>

      static void Print(); // will print "T(*)[]"
      ;


      And when in our code we will do something like this:



      int (*arrptr)[10];
      CoolClass<decltype(arrptr)>::Print();


      The console will print "T(*)[sized]" as we expected (considering that we wrote implementation for all methods of course). But what if we will wrote our code like this:



      int (**arrptr_d)[10];
      CoolClass<decltype(arrptr_d)>::Print();


      In this case the console will actually print "Default". We could write another specialization like this:



      template<typename T, size_t S>
      struct CoolStruct<T(**)[S]>

      static void Print(); // will print "T(*)[sized]"
      ;


      but I want to find another solution (I don't want to write another dozen specializations when I would need to use int(****)[]). So considering we using C++17 standard, is there a way to do such a thing?



      P.S. And yes, sorry for my terrible English; it is not my native language.







      c++ templates c++17






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 26 at 12:03









      Jon Harper

      3,02621130




      3,02621130










      asked Mar 22 at 15:50









      VladVlad

      92




      92






















          1 Answer
          1






          active

          oldest

          votes


















          1














          If you don't care about other pointers, you could do a partial specialization that delegates double pointers to the specialization for single pointers:



          template<typename Pointee>
          struct CoolStruct<Pointee**>
          static void Print()
          CoolStruct<Pointee*>::Print();

          ;





          share|improve this answer























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            1 Answer
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            active

            oldest

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            active

            oldest

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            1














            If you don't care about other pointers, you could do a partial specialization that delegates double pointers to the specialization for single pointers:



            template<typename Pointee>
            struct CoolStruct<Pointee**>
            static void Print()
            CoolStruct<Pointee*>::Print();

            ;





            share|improve this answer



























              1














              If you don't care about other pointers, you could do a partial specialization that delegates double pointers to the specialization for single pointers:



              template<typename Pointee>
              struct CoolStruct<Pointee**>
              static void Print()
              CoolStruct<Pointee*>::Print();

              ;





              share|improve this answer

























                1












                1








                1







                If you don't care about other pointers, you could do a partial specialization that delegates double pointers to the specialization for single pointers:



                template<typename Pointee>
                struct CoolStruct<Pointee**>
                static void Print()
                CoolStruct<Pointee*>::Print();

                ;





                share|improve this answer













                If you don't care about other pointers, you could do a partial specialization that delegates double pointers to the specialization for single pointers:



                template<typename Pointee>
                struct CoolStruct<Pointee**>
                static void Print()
                CoolStruct<Pointee*>::Print();

                ;






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Mar 26 at 9:28









                Anthony WilliamsAnthony Williams

                52.8k9102140




                52.8k9102140





























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