How can specializations of a template class be mutual friends ?

[Ulf Rehmann]

I would like to make two specializations of the same template 
mutual friends. However:

The following code gives an error under g++ 2.95.2:

template<class Scalar> 
class complex{
  Scalar re, im;
public:
  complex(Scalar r, Scalar i) { re=r, im=i;};
  template<class T>
  complex(const complex<T>& c) { re=c.re; im=c.im; }; 
  friend class complex<float>;
  friend class complex<double>;
};
int main() {
  complex<float>  cf(0,0);
  complex<double> cd = cf;
}

This is the error message:


g++ ttt.cc
ttt.cc: In method `complex<double>::complex<float>(const complex<float> &)':
ttt.cc:13:   instantiated from here
ttt.cc:3: `float complex<float>::re' is private
ttt.cc:7: within this context
ttt.cc:3: `float complex<float>::im' is private
ttt.cc:7: within this context

What is wrong?

Any hint is welcome.

Ulf Rehmann