Default template parameter with class












11















I've just found out about a strange syntax for default template parameters



template<class T = class Z>
struct X
{};


What does the second "class" keyword mean in this context?










share|improve this question



























    11















    I've just found out about a strange syntax for default template parameters



    template<class T = class Z>
    struct X
    {};


    What does the second "class" keyword mean in this context?










    share|improve this question

























      11












      11








      11


      2






      I've just found out about a strange syntax for default template parameters



      template<class T = class Z>
      struct X
      {};


      What does the second "class" keyword mean in this context?










      share|improve this question














      I've just found out about a strange syntax for default template parameters



      template<class T = class Z>
      struct X
      {};


      What does the second "class" keyword mean in this context?







      c++ templates default-value






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Jan 27 at 13:02









      DunsDuns

      905




      905
























          1 Answer
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          13














          It's nothing special really. C++ allows you to refer to a class via an elaborated type specifier. E.g.



          void foo(class bar*);


          This declares a function foo that accepts an argument of the type bar*. If bar was not declared previously, this elaborate type specifier constitutes a declaration of bar in the namespace containing foo. I.e. as if you had written:



          class bar;
          void foo(bar*);


          Back to your example, X is a class template that expects a single type parameter, denoted by class T, but could have been denoted just the same as typename T. Said type parameter has a default argument, named by the elaborated class specifier class Z. That declaration can be rewritten just like the function above:



          class Z;
          template<class T = Z>
          struct X
          {};





          share|improve this answer
























          • great explanation, thank you very much

            – Duns
            Jan 27 at 13:23











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          13














          It's nothing special really. C++ allows you to refer to a class via an elaborated type specifier. E.g.



          void foo(class bar*);


          This declares a function foo that accepts an argument of the type bar*. If bar was not declared previously, this elaborate type specifier constitutes a declaration of bar in the namespace containing foo. I.e. as if you had written:



          class bar;
          void foo(bar*);


          Back to your example, X is a class template that expects a single type parameter, denoted by class T, but could have been denoted just the same as typename T. Said type parameter has a default argument, named by the elaborated class specifier class Z. That declaration can be rewritten just like the function above:



          class Z;
          template<class T = Z>
          struct X
          {};





          share|improve this answer
























          • great explanation, thank you very much

            – Duns
            Jan 27 at 13:23
















          13














          It's nothing special really. C++ allows you to refer to a class via an elaborated type specifier. E.g.



          void foo(class bar*);


          This declares a function foo that accepts an argument of the type bar*. If bar was not declared previously, this elaborate type specifier constitutes a declaration of bar in the namespace containing foo. I.e. as if you had written:



          class bar;
          void foo(bar*);


          Back to your example, X is a class template that expects a single type parameter, denoted by class T, but could have been denoted just the same as typename T. Said type parameter has a default argument, named by the elaborated class specifier class Z. That declaration can be rewritten just like the function above:



          class Z;
          template<class T = Z>
          struct X
          {};





          share|improve this answer
























          • great explanation, thank you very much

            – Duns
            Jan 27 at 13:23














          13












          13








          13







          It's nothing special really. C++ allows you to refer to a class via an elaborated type specifier. E.g.



          void foo(class bar*);


          This declares a function foo that accepts an argument of the type bar*. If bar was not declared previously, this elaborate type specifier constitutes a declaration of bar in the namespace containing foo. I.e. as if you had written:



          class bar;
          void foo(bar*);


          Back to your example, X is a class template that expects a single type parameter, denoted by class T, but could have been denoted just the same as typename T. Said type parameter has a default argument, named by the elaborated class specifier class Z. That declaration can be rewritten just like the function above:



          class Z;
          template<class T = Z>
          struct X
          {};





          share|improve this answer













          It's nothing special really. C++ allows you to refer to a class via an elaborated type specifier. E.g.



          void foo(class bar*);


          This declares a function foo that accepts an argument of the type bar*. If bar was not declared previously, this elaborate type specifier constitutes a declaration of bar in the namespace containing foo. I.e. as if you had written:



          class bar;
          void foo(bar*);


          Back to your example, X is a class template that expects a single type parameter, denoted by class T, but could have been denoted just the same as typename T. Said type parameter has a default argument, named by the elaborated class specifier class Z. That declaration can be rewritten just like the function above:



          class Z;
          template<class T = Z>
          struct X
          {};






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jan 27 at 13:08









          StoryTellerStoryTeller

          100k12204273




          100k12204273













          • great explanation, thank you very much

            – Duns
            Jan 27 at 13:23



















          • great explanation, thank you very much

            – Duns
            Jan 27 at 13:23

















          great explanation, thank you very much

          – Duns
          Jan 27 at 13:23





          great explanation, thank you very much

          – Duns
          Jan 27 at 13:23




















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