Re: Dependent type in template base class

From:

"=?iso-8859-1?q?Daniel_Kr=FCgler?=" <daniel.kruegler@googlemail.com>

Newsgroups:

comp.lang.c++.moderated

Date:

12 Jan 2007 09:51:17 -0500

Message-ID:

<1168590235.984363.3810@q2g2000cwa.googlegroups.com>

Stephan Tolksdorf schrieb:

I'd like to make the name of a member type in a template base class
directly available to the child class. (I don't want to qualify each
reference to the type with "typename Base<T>::" in order to avoid
extremely ugly code).
I'm currently doing this with an using declaration as in the following
code snippet:

template <typename T>
class base {

// Public or protected access needed:
public:

     struct param_type { };
     void f(param_type& p) { }
};

template <typename T>
class test : base<T> {
     using typename base<T>::param_type;
     test(param_type& p) { } // GCC doesn't recognize param_type
}

While Comeau and Visual Studio don't complain about this code, GCC
doesn't seem to grok it.

Could I use a typedef instead of the using declaration as a workaround,
or would that entail other problems?

Normally there should be no problems. The most defensive approach
would be to use the same name for the typedef as in the base class
template, i.e.

template <typename T>
class test : base<T> {
public:
      typedef typename base<T>::param_type param_type;
      test(param_type& p) { }
};

Greetings from Bremen,

Daniel Kr|gler

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