Rationale (September, 2012):

The concerns in this issue were addressed by the resolutions of issues 403 and 557.

The last bullet of the second paragraph of section 6.5.4 [basic.lookup.argdep] says that:

If T is a template-id, its associated namespaces and classes are the namespace in which the template is defined; for member templates, the member template's class; the namespaces and classes associated with the types of the template arguments provided for template type parameters (excluding template template parameters); the namespaces in which any template template arguments are defined; and the classes in which any member templates used as template template arguments are defined.

The first problem with this wording is that it is misleading, since one cannot get such a function argument whose type would be a template-id. The bullet should be speaking about template specializations instead.

The second problem is owing to the use of the word "defined" in the phrases "are the namespace in which the template is defined", "in which any template template arguments are defined", and "as template template arguments are defined". The bullet should use the word "declared" instead, since scenarios like the one below are possible:

namespace  A  {

   template<class T>
   struct  test  {

      template<class U>
      struct  mem_templ  {   };

   };

   // declaration in namespace 'A'
   template<> template<>
   struct  test<int>::mem_templ<int>;

   void  foo(test<int>::mem_templ<int>&)
   {   }

}

// definition in the global namespace
template<> template<>
struct  A::test<int>::mem_templ<int>  {
};

int  main()
{
   A::test<int>::mem_templ<int>   inst;
   // According to the current definition of 3.4.2
   // foo is not found.
   foo(inst);
}

In addition, the bullet doesn't make it clear whether a T which is a class template specialization must also be treated as a class type, i.e. if the contents of the second bullet of the second paragraph of section 6.5.4 [basic.lookup.argdep].

• If T is a class type (including unions), its associated classes are: the class itself; the class of which it is a member, if any; and its direct and indirect base classes. Its associated namespaces are the namespaces in which its associated classes are defined. [This wording is as updated by core issue 90.]

template<class T>
struct  slist_iterator  {
   friend bool  operator==(const slist_iterator& x, const slist_iterator& y)
   {   return  true;   }
};

template<class T>
struct  slist  {
   typedef slist_iterator<T>   iterator;
   iterator  begin()
   {   return  iterator();   }
   iterator  end()
   {   return  iterator();   }
};

int  main()
{
   slist<int>   my_list;
   slist<int>::iterator   mi1 = my_list.begin(),  mi2 = my_list.end();
   // Must the the friend function declaration
   // bool  operator==(const slist_iterator<int>&, const slist_iterator<int>&);
   // be found through argument dependent lookup? I.e. is the specialization
   // 'slist<int>' the associated class of the arguments 'mi1' and 'mi2'. If we
   // apply only the contents of the last bullet of 3.4.2/2, then the type
   // 'slist_iterator<int>' has no associated classes and the friend declaration
   // is not found.
   mi1 == mi2;
}

Suggested resolution:

Replace the last bullet of the second paragraph of section 6.5.4 [basic.lookup.argdep]

• If T is a template-id, its associated namespaces and classes are the namespace in which the template is defined; for member templates, the member template's class; the namespaces and classes associated with the types of the template arguments provided for template type parameters (excluding template template parameters); the namespaces in which any template template arguments are defined; and the classes in which any member templates used as template template arguments are defined.

• If T is a class template specialization, its associated namespaces and classes are those associated with T when T is regarded as a class type; the namespaces and classes associated with the types of the template arguments provided for template type parameters (excluding template template parameters); the namespaces in which the primary templates making template template arguments are declared; and the classes in which any primary member templates used as template template arguments are declared.
• If T is a function template specialization, its associated namespaces and classes are those associated with T when T is regarded as a function type; the namespaces and classes associated with the types of the template arguments provided for template type parameters (excluding template template parameters); the namespaces in which the primary templates making template template arguments are declared; and the classes in which any primary member templates used as template template arguments are declared.

Replace the second bullet of the second paragraph of section 6.5.4 [basic.lookup.argdep]

• If T is a class type (including unions), its associated classes are: the class itself; the class of which it is a member, if any; and its direct and indirect base classes. Its associated namespaces are the namespaces in which its associated classes are defined.

• If T is a class type (including unions), its associated classes are: the class itself; the class of which it is a member, if any; and its direct and indirect base classes. Its associated namespaces are the namespaces in which its associated classes are declared [Note: in case of any of the associated classes being a class template specialization, its associated namespace is acually the namespace containing the declaration of the primary class template of the class template specialization].