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[QUOTE="Kai-Uwe Bux, post: 2573153"]
[snip]

Maybe someting like this (although it maybe overly complicated):

#include <string>
#include <iostream>

struct a_string : private std::string {

a_string ( void )
  {}
  
  template < typename A >
  a_string ( A const & a )
    : std::string( a )
  {}

template < typename A, typename B >
  a_string ( A const & a, B const & b )
    : std::string( a, b )
  {}

a_string & operator+= ( a_string const & other ) {
    std::string::operator+=( other );
    return ( *this );
  }

std::string const & str ( void ) const {
    return ( *this );
  }
  
};

struct b_string : private std::string {

b_string ( void )
  {}
  
  template < typename A >
  b_string ( A const & a )
    : std::string( a )
  {}

template < typename A, typename B >
  b_string ( A const & a, B const & b )
    : std::string( a, b )
  {}
  
  b_string & operator+= ( b_string const & other ) {
    std::string::operator+=( other );
    return ( *this );
  }
  
  std::string const & str ( void ) const {
    return ( *this );
  }

};

struct new_line_const {

new_line_const ( int )
  {}
  
  operator a_string ( void ) const {
    return ( a_string( "hello" ) );
  }

operator b_string ( void ) const {
    return ( b_string( "world" ) );
  }

};

static
new_line_const const new_line = 0;

int main ( void ) {
  a_string a ( "greetings: " );
  a += new_line;
  std::cout << a.str() << '\n';
  
  b_string b ( "greetings: " );
  b += new_line;
  std::cout << b.str() << '\n';
}

Also note that the compiler will perform only one user defined conversion,
so this trick might not work in all circumstances.

Best

Kai-Uwe Bux
[/QUOTE]

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