Re: Templates, inheritance and variable visibility

From:

"Alf P. Steinbach" <alfps@start.no>

Newsgroups:

comp.lang.c++

Date:

Sat, 04 Apr 2009 02:33:28 +0200

Message-ID:

<gr6a00$bt3$1@news.motzarella.org>

* ajhietanen@gmail.com:

Hi everybody,

I have a following class structure, where B inherits A and specifies
the other template parameter.

template <typename T,typename V>
class A {
protected:
T a_;
};

template <typename V>
class B: public A<int,V>{
public:
B(){a_=2;} // <-- error: ?a_? was not declared in this scope
};

The problem is that compiler does not find the declaration of a_. The
program compiles if I write
A<int,V>::a_ = 2;
or I specify both of the template parameters. The compiler is g++
(gcc) 4.2.4

Why is this?

The definition or existence of 'a_' depends on template parameter.

You have to tell the compiler in some way, by qualification or 'using', that it
comes from the base class and is not from an enclosing scope.

Exactly why that rule was chosen instead of the for C++98 more reasonable of
having to qualify use of enclosing scope names is a mystery.

And is there a way to use the variable a_ without the
whole specification, e.g., with keyword using?

Yes, exactly.

Cheers & hth.,

- Alf

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