Re: templates - method return value is multiple types

Christopher <cpisz@austin.rr.com> writes:


  You can ?return? a value of either ?int? or ?double? type in a
  certain figurative sense:

  I call ?callee?, which will ?return? either an ?int? or a ?double?:

callee( this );
::std::cout << num->type() << '\n';
::std::cout << num->text() << '\n';

  . This will print either

int
1

  or

double
1

  . The callee is looking like:

void callee( visitor * const v )
{ if( ::std::rand() )v->accept( 1 );
  else v->accept( 1.0 ); }

  , instead of an actual return, he uses ?v->accept?.

  Now, the complete program:

#include <iostream>
#include <ostream>
#include <cstdlib>
#include <string>
#include <sstream>

struct visitor
{ virtual void accept( int const value )= 0;
  virtual void accept( double const value )= 0; };

struct number
{ virtual ::std::string type()= 0;
  virtual ::std::string text()= 0; };

struct integer : public number
{ int val;
  integer( int const n ): val( n ){};
  virtual ::std::string type(){ return "int"; };
  virtual ::std::string text(){ std::stringstream out;
    out << val; return out.str(); }};

struct floating : public number
{ double val;
  floating( double const n ): val( n ){};
  virtual ::std::string type(){ return "double"; };
  virtual ::std::string text(){ std::stringstream out;
    out << val; return out.str(); }};

void callee( visitor * const v )
{ if( ::std::rand() )v->accept( 1 );
  else v->accept( 1.0 ); }

struct caller : public visitor
{ number * num;
  void accept( int const value ){ this->num = new integer( value ); };
  void accept( double const value ){ this->num = new floating( value ); };
  void example()
  { callee( this );
    ::std::cout << num->type() << '\n';
    ::std::cout << num->text() << '\n'; }};

int main(){ caller().example(); }