template friend for operator- How to

From:
Helmut Jarausch <jarausch@igpm.rwth-aachen.de>
Newsgroups:
comp.lang.c++
Date:
8 Jul 2014 09:46:36 GMT
Message-ID:
<c21svsFq7jlU1@mid.dfncis.de>
Hi,

I have a problem with a friend declaration of the dyadic operator- IFF there is a monadic operator-, as well.

Note that the friend declaration below is accepted if and only if the definition of the monadic operator
is removed.
Many thanks for a hint,
Helmut

template <typename INT,INT P> class Zp;

template <typename INT,INT P>
Zp<INT,P> operator-(const Zp<INT,P>& a, const Zp<INT,P>& b);

template <typename INT,INT P>

class Zp {
public:
  static const INT p = P;
private:

  INT val;

public:

  Zp() : val(0) {}
  Zp( INT x ) : val(x%p) { if (x < 0 ) x+= p; }
  
  Zp operator-() const { return Zp(p-val); } // if this is deleted, it works <<<<<
  friend Zp<INT,P> operator- <>(const Zp<INT,P>& a, const Zp<INT,P>& b);

/* gcc-4.9.0 says
Quest_Templ.C:22:28: error: declaration of 'operator-' as non-function
   friend Zp<INT,P> operator- <>(const Zp<INT,P>& a, const Zp<INT,P>& b);
                            ^
Quest_Templ.C:22:28: error: expected ';' at end of member declaration
Quest_Templ.C:22:30: error: expected unqualified-id before '<' token
   friend Zp<INT,P> operator- <>(const Zp<INT,P>& a, const Zp<INT,P>& b);
*/
};

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