C ++中的专用朋友功能

Question

I have a function template foo that has to perform different computations depending on whether the template parameter is a real or a complex number. In any case, the result will be a real number, eg double , even if the template parameter is std::complex<double> . Therefore, my function looks as follows:

template <class S>
struct RealType
{
  typedef S type;
};

template <class S>
struct RealType<std::complex<S> >
{
  typedef S type;
};

template <class S>
class C;

template <class S>
typename RealType<S>::type foo(C<S> &c);

template <class S>
typename RealType<std::complex<S> >::type foo(C<std::complex<S> > &c);

Now foo must be a friend function of the class C , so I made the following declaration:

template <class S>
class C
{
  friend typename RealType<S>::type foo(C<S> &c);
  // ...
};

However, when I instantiate C<std::complex<double> > the compiler says that foo cannot access the private members of c . It works fine for C<double> . Is there any solution to this (that works with C++98)? I understand that foo cannot be a member of C , as this will prevent the partial specialization.

BTW: Is this really a specialization? The signatures of both versions of foo look the same, but in fact they are somewhat different when the real types are plugged in.

Answer 1

Make sure that class C declares foo a friend after foo has been declared.

You may need to use forward declarations for that:

template <class S> class C;

// foo declarations here

template <class S> class C 
{ 
    template<class X> friend typename RealType<X>::type foo(C<X> &c); // friend the template
    friend typename RealType<S>::type foo<>(C &c); // friend the specialization
};