Hello, for formatting output I wrote the following classes: ---8<--- namespace /* anonymous */ { namespace mpl = ::boost::mpl; using ::boost::enable_if; using ::boost::xpressive::sregex; using ::boost::xpressive::as_xpr; template<typename StringT> struct is_string : mpl::false_ { }; template<> struct is_string<std::string> : mpl::true_ { }; template <typename Tp, typename Enable = void> struct formatter { formatter( const Tp& data ) : output( data ) { } std::ostream& write_on( std::ostream& os ) const { os << output; return os; } private: const Tp& output; }; template <typename Tp> struct formatter<Tp, typename enable_if< is_string<Tp> >::type> { formatter( const Tp& data ) : output( data ), re( as_xpr("\r") ), format("\n") { } std::ostream& write_on( std::ostream& os ) const { using ::boost::xpressive::regex_replace; std::ostream_iterator<std::ostream::char_type> out_iter( os ); regex_replace( out_iter, output.begin(), output.end(), re, format ); return os; } private: const Tp& output; const sregex re; const std::string format; }; } --->8--- Mainly, this replaces all '\r' by '\n' in all string in the given data, e.g.: ---8<--- template<typename DataT> void foo::write_key_data_pair( std::ostream& os, const std::string& key, const DataT& data ) const { os << "[" << key << "]" << " " ; formatter<DataT>( data ).write_on( os ); os << std::endl ; } --->8--- Anyway, overloading global operator<<() as template<typename T> static inline std::ostream& operator<<( std::ostream& os, const T& rhs ) { return rhs.write_on( os ); } would force all to have a write_on member function, isn't it. Is there a way around and even use operator<<() notation ?? Thanks, Olaf