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From: fuchs@it.uka.de (Harald Fuchs)
Newsgroups: comp.lang.c++
Subject: Re: How smart is operator resolution?
Message-ID: <fuchs.659902282@t500m0>
Date: 29 Nov 90 18:11:22 GMT
References: <1747@umriscc.isc.umr.edu>
Sender: news@ira.uka.de (USENET News System)
Organization: University of Karlsruhe, FRG
Lines: 49

jamesh@cs.umr.edu (James Hartley) writes:

>I recently tried taking advantage of virtual functions by creating a generic
>routine which would take a base class pointer and output the appropriate 
>derived class where the insertor was properly overloaded.  The following 
>program attempts to illustrate the idea.  Turbo C++ balks at the output
>statement in main() because the insertor is not defined with respect to
>the base class.  Is this operator resolution beyond the restraints of the
>language, or is this a bug in the compiler?

You get the inserter you want by making it virtual. You can't do this
directly because the standard insertion operator<< expects your
"figure" to be the 2nd argument. Do it like that:

class figure {                  // abstract base class definition
protected:
    float r, h;
public:
    virtual float volume(void) = 0;
    virtual void printOn (ostream&) const = 0;
};

inline ostream& operator<< (ostream& s, const figure& x) {
  x.printOn (s);
  return s;
}

class sphere : public figure {  // derived class definition
public:
    sphere(float radius)  { r = radius; }
    virtual float volume(void)  { return 4.0 / 3.0 * pi * r * r * r; }
    void printOn (ostream&) const;
};

void sphere::printOn (ostream& strm) const {
    strm << "VOLUME OF SPHERE:\n";
    strm << "radius = " << sph.r << "\n";
    strm << "volume = " << sph.volume() << "\n";
}

main() {
    figure *fig = new sphere (2.0);
    cout << *fig << "\n";
    return 0;
}
--

Harald Fuchs <fuchs@it.uka.de> <fuchs%it.uka.de@relay.cs.net> ...
<fuchs@telematik.informatik.uni-karlsruhe.dbp.de>   *gulp*