Re: Accessing types in base class?

From:
Pete Becker <pete@versatilecoding.com>
Newsgroups:
comp.lang.c++
Date:
Sat, 27 Mar 2010 09:23:44 -0400
Message-ID:
<aMWdnfyiwq99mzPWnZ2dnUVZ_uCdnZ2d@giganews.com>
none wrote:

I have this base class:

template<typename J>
class Processor {
public:
  const unsigned int Dimension = J::Dimension;
  typedef typename J::ImageType ImageType;

  Processor ();
  virtual ~Processor ();

};
#endif

In a sub class I need to use the types from the base class:

  template <typename J>
  class SubProcessor : public Processor<J>{
  public:

    void test(ImageType image) {
       // do stuff

    }

};

But that is not possible! The base class type "ImageType" is not visible
in the sub class. In the sub class I could just do the same as in the
base class:

  typedef typename J::ImageType ImageType;

but what is the point of using inheritance then?


The rules for deriving from template specializations are a bit more
complicated, because of the possibility of explicitly specializing a
template. There's no requirement, for example, that Processor<int> have
a member named ImageType, because it could be an explicit specialization:

template <> Processor<int> { /* empty definition */ };

So if you want to talk about a member of a base class that's a template
specialization you have to add qualifiers to the names the members that
you want, generally with this-> to talk about a member function or a
data member, or Processor<J>:: to talk about a type that's named in the
template. Once you've done that, trying to use an explicit
specialization that doesn't define the member results in an error.

template <typename J>
class SubProcessor : public Processor<J> {
public:
   void test(Processor<J>::ImageType image) { } // OK
};

SubProcessor<double> spd; // OK
SubProcessor<int> ipd; // error: no ImageType in Processor<int>

--
   Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of
"The Standard C++ Library Extensions: a Tutorial and Reference"
(www.petebecker.com/tr1book)

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