Question

I have a class that uses a nested class, and want to use the nested class operator<< to define operator<< in the upper class. Here is how my code looks like:

#include <memory>
#include <iostream>

template<typename T>
struct classA {
  struct classB
  {
    template<typename U>
    friend inline std::ostream& operator<< (std::ostream &out,
                                            const typename classA<U>::classB &b);
  };

  classB root;

  template<typename U>
  friend std::ostream& operator<< (std::ostream &out,
                                   const classA<U> &tree);
};

template<typename T>
inline std::ostream& operator<< (std::ostream &out,
                                 const classA<T> &tree)
{
  out << tree.root;
  return out;
}

template<typename T>
inline std::ostream& operator<< (std::ostream &out,
                                 const typename classA<T>::classB &b)
{
  return out;
}

int main()
{
  classA<int> a;
  std::cout << a;
}

• When compiling without support for C++11, the definition of operator<< for the inner class seems not to be found by the compiler:

  so.hpp:24:7: error: no match for ‘operator<<’ in ‘out << tree.classA<int>::root’
  so.hpp:24:7: note: candidates are: ...
  

• With GCC 4.6 and 4.7 when compiling with std=c++0x:

  so.hpp:21:3: error: cannot bind ‘std::ostream {aka std::basic_ostream<char>}’ lvalue to ‘std::basic_ostream<char>&&’
  In file included from /usr/include/c++/4.7/iostream:40:0,
                   from so.hpp:2:
  /usr/include/c++/4.7/ostream:600:5: error:   initializing argument 1 of ‘std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&&, const _Tp&) [with _CharT = char; _Traits = std::char_traits<char>; _Tp = classA<int>::classB]’
  

Can someone tell me why this code is not legal, and what’s the best way to do what I want?

Answer 1

Bo provided the reason why this is happening (the type T is not deducible in the call to the nested operator<<. A simple workaround for this, and something that I recommend in general, not only here, is not befriending a template, but rather a single free function. For that you will need to define the function inline:

template<typename T>
struct classA {
  struct classB
  {
    friend inline std::ostream& operator<< (std::ostream &out,
                                            const classB &b) {
       // definition goes here
    }
  };

  classB root;

  friend std::ostream& operator<< (std::ostream &out,
                                   const classA<U> &tree) {
       // definition goes here
  }
};

There are a couple of differences among the two approaches. The most important one is that this approach will have the compiler define a non-templated overload for operator<< for each instantiation of the template, which because it is no longer a template, does not depend on deducing the arguments. Another side effects are that the approach is a little tighter (you are only befriending one function, while in your initial approach you befriended the template and all possible instantiations (which can be used as a loophole to gain access to your class internals). Finally the functions so defined will only be found through ADL, so there are less overloads of operator<< for the compiler to consider when the argument is not ClassA<T> or ClassA<T>::ClassB.

---

How access can be gained with your approach

namespace {
   struct intruder {
       ClassA & ref;
       intruder( ClassA& r ) : ref(r) {}
   };
   template <>
   std::ostream& operator<< <intruder>( std::ostream& _, ClassA<intruder> const& i ) {
       std::cout << i.ref.private_member << std::endl;
       return _;
   }
}

Alternative

Alternatively you can befriend a particular specialization of a template. That will solve the intruder problem, as it will only be open to operator<< to ClassA<intruder>, which has a much lesser impact. But this will not solve your particular issue, as the type would still not be deducible.