Path: utzoo!censor!geac!torsqnt!news-server.csri.toronto.edu!clyde.concordia.ca!uunet!microsoft!jimad
From: jimad@microsoft.UUCP (Jim ADCOCK)
Newsgroups: comp.object
Subject: Re: inheritance and `type loss'
Message-ID: <55775@microsoft.UUCP>
Date: 11 Jul 90 18:16:17 GMT
References: <SRA.90Jul10155606@alonzo.ecs.soton.ac.uk>
Reply-To: jimad@microsoft.UUCP (Jim ADCOCK)
Distribution: comp
Organization: Microsoft Corp., Redmond WA
Lines: 57

In article <SRA.90Jul10155606@alonzo.ecs.soton.ac.uk> sra@ecs.soton.ac.uk (Stephen Adams) writes:

>(a)  what should   (V+V).image  yield?  "10" or "X"?
>		   (V+1).image          "6" or "VI"?
>		   (1+V).image          "6" or "VI" or error (in +)?
>
>(b)  what do various object oriented systems actually do?
>

C++ provides are variety of ways to allow a programmer to either return
an int or a roman in the above example.  Generalizing the problem to
full polymorphic behavior is quite doable -- but ugly, requiring a double
dispatch.  Say in general, one wants:

	aDog = aDog + aDog;

but in specific one wants:

	aPoodle = aPoodleF + aPoodleM;
	aCockerSpaniel = aCockerSpanielF + aCockerSpanielM;
	aCockapoo = aPoodleF + aCockerSpanielM;
	aCockapoo = aCockerSpanielF + aPoodleM;

--So one can end up specifying a lot of flavors of "+" to get exactly the
behavior one wants.  [I'm using inter-breeding as a metaphor -- I'm not
suggesting the above is a good overloading of "+"]

--- 

void image(int i) { printf("%d\n", i); }

class roman1
{
protected: int i;
public:
    roman1(int ii) : i(ii) {}
    operator int() { return i; }
// a simplified version of roman numerals -- works for small positive ints :-)
    friend void image(roman1 r) { while (r.i--) printf("I"); printf("\n"); }
};

class roman2 : public roman1
{
public:
    roman2(int i) : roman1(i) {}
    friend roman2 operator+(int i, roman2 r2) { return i + r2.i; }
};

main()
{
    int anInt = 1;
    roman1 aRoman1 = 2;
    image(anInt + aRoman1); // prints "3"
    roman2 aRoman2 = 2;
    image(anInt + aRoman2); // prints "III"
    return 0;
}