Re: What is necessary to make a template's method dependent

From:

=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <daniel.kruegler@googlemail.com>

Newsgroups:

comp.lang.c++.moderated

Date:

Mon, 22 Jul 2013 10:17:08 -0700 (PDT)

Message-ID:

<ksjhci$3v0$1@dont-email.me>

On 2013-07-22 04:57, heideggm@googlemail.com wrote:

let's say, we have the following class templates:

template<class T> class MyClass
{
         public:

         void myMethod(T* myParam)
         {
             std::cout<<"Whatever."<<endl;
         }
};

template<class T> class DerivedClass : public MyClass<T>
{
     public:

         void myOtherMethod(T* myParam)
         {
                 myMethod(myParam);
         }
};

Now I create an instance of DerivedClass and call myOtherMethod from it.
The compiler complaints that he cannot find myMethod in myOtherMethod
and wants me to use the this pointer.

My question is shouldn't the two phase name lookup recognize that
myMethod depends on the template argument (via T* myParam) and therefore
find it during the second phase?

No, it shouldn't. The language is very clear in 14.6.2 [temp.dep] p3:

"In the definition of a class or class template, if a base class depends
on a template-parameter, the base class scope is not examined during
unqualified name lookup either at the point of definition of the class
template or member or during an instantiation of the class template or
member."

Note that the wording does not distinguish between dependent and
non-dependent expressions in this context. The relevant point is that a
dependent base-classes stops unqualified name lookup.

HTH & Greetings from Bremen,

Daniel Kr?gler

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