TITLE: how to use "inherited" typedefs

(Newsgroups: comp.lang.c++.moderated, 21 Feb 2000)

MULCHANDANI: Kishore Mulchandani <mool@mediaone.net>

>template <class T>
>class A
>{
> public :
>     typedef T* ptrT;
>     A(ptrT a){e_a = a;}
>
>     private:
>         ptrT a;
>
>};
>
>template <class T>
>class B: public A<T>
>{
>     public :
>         B(ptrT t) : A( t) {} ////Error: illegal constructor/destructor
declaration
>};

NARAN: sbnaran@uiuc.edu (Siemel B. Naran)

Your Metroworks compiler is right.  One may specialize A<T> so that it
does not have the 'ptrT' typedef, or so that 'ptrT' is a static const
variable.  Then the constructor of B would be an error.

But the rules of C++ ensure that template code is processed and
precompiled as much as possible.  This means it should be known
exactly where 'ptrT' comes from.  Hence, your two solutions are,

 template <class T>
 class B: public A<T>
 {
      public :
          using A<T>::ptrT; // or "typdef typename A<T>::ptrT ptrT"
          B(ptrT t) : A( t) {}
 };

 template <class T>
 class B: public A<T>
 {
      public :
          B(typename A<T>::ptrT t) : A( t) {}
 };


But I'd just use "T *" once and for all.



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