Re: Template friend in templat class

I have a template class:

template <typename Base_t> class Rational_t {

// Definition of private and public members left out.
// Probably not relevant.

};

I want an output operator << for this class for different output =

I am not sure that I completely understand the hierarchy of the i/o =

classes, but I have the impression that the base class in which the
operator << is defined is a template class. (It would be much easier if =

first class in which the << operator is defined, were a non-template
class.) So, I need at least two template parameters for my << operator. =

Therefore, in the same header file I had:

template <typename Base_t, typename ostream_t>
ostream_t & operator<<
 (ostream_t & out, const Rational_t<Base_t> & Rat) {

// Implementation left out.
// Probably not relevant.

}

This works.

Now, I want to change the implementation of the << operator a bit, for
which it needs access to some private members of the Rational_t class.
Therefore, I want to make the << operator a friend of the class. This
requires a few forward declarations at the start of the file.

template <typename Base_t> class Rational_t;

template <typename Base_t, typename ostream_t>
ostream_t & operator<<
 (ostream_t & out, const Rational_t<Base_t> & Rat);

This also still works, as long as the implementation of the << operator =

does not access the private members of Rational_t.

I have tried several ways to write the friend declaration, but non of =

works.
I think the last one I tried was:

template <typename Base_t> class Rational_t {

template <typename ostream_t>
 friend ostream_t& operator<<
  (ostream_t& out, const Rational_t & Rat);

// Definition of private and public members left out.
// Probably not relevant.

};

When I use g++ under Linux
(g++ (SUSE Linux) 4.3.4 [gcc-4_3-branch revision 152973]),
I see a lot of messages like (One long line wrapped by me.):

TestRational.cpp:(.text+0xa93): undefined reference to
`std::basic_ostream<char, std::char_traits<char> >&
Rational::operator<<
<std::basic_ostream<char, std::char_traits<char> > >
(std::basic_ostream<char, std::char_traits<char> >&,
Rational::Rational_t<int> const&)'

It seems that the compiler has no problem with the code, but the linker =

has.
Usually this indicates that the friend declaration is interpreted as a
forward declaration, for which no implementation is found. So =

something is wrong with the friend declaration. But I don't see how I =

make it work.
Any suggestion?