Re: please help with friend function in nested class template

From:
"Igor Tandetnik" <itandetnik@mvps.org>
Newsgroups:
microsoft.public.vc.language
Date:
Thu, 10 Sep 2009 17:46:42 -0400
Message-ID:
<e$fRPBmMKHA.4780@TK2MSFTNGP05.phx.gbl>
LiDongning <lidongning@gmail.com> wrote:

I want to declare a function template as a friend in a nested class
template. The code is as follows.

template <class T>
class a1
{
public:
     class a2
     {
     T att;
     template <class T> friend const typename a1<T>::a2 operator +
(const typename a1<T>::a2&, const typename a1<T>::a2&);
     };
};

template <class T>
const typename a1<T>::a2 operator + (const typename a1<T>::a2&, const
typename a1<T>::a2&)
{
     return a1<T>::a2 xx;
}

int main()
{
   a1<int> x;
   a1<int>::a2 y;
   y + y; //error! c2783
   system("PAUSE");
   return 0;
}

Complied in VC2008. Got error C2783.


Friend declaration is a red herring. The real problem is that, in
operator+ invocation, T is in a non-deducible context. Consider:

template <typename T>
struct Outer {
  struct Inner {};
};

template <typename T>
void f(const typename Outer<T>::Inner& x) {}

int main() {
  Outer<int>::Inner y;
  f(y); // error here
  return 0;
}

In f(y) call, the compiler cannot deduce that T==int. For all it knows,
you have something like this somewhere in your program:

template<>
struct Outer<long> {
  typedef Outer<int>::Inner Inner;
};

Now Outer<int>::Inner and Outer<long>::Inner both name the same type.
--
With best wishes,
    Igor Tandetnik

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