Path: utzoo!mnetor!uunet!husc6!mit-eddie!uw-beaver!tikal!hplsla!jima
From: jima@hplsla.HP.COM (jim adcock )
Newsgroups: comp.lang.c
Subject: Re: why just a power operator?
Message-ID: <5260002@hplsla.HP.COM>
Date: 11 Jan 88 21:05:25 GMT
References: <3410@megaron.arizona.edu>
Organization: HP Lake Stevens, WA
Lines: 74


/*

Regards: overloading "->" for "pow"

Sorry, I should have known better than to try to
sell "software futures."

Overloading "->" MIGHT have been "nice" since
it has about the precedence and binding
required for a good power operator.

Unfortunately, as Bjarne points out, there are
three problems with this: 1) "->" expects a 
member on its right hand side -- you don't get
to "overload" C++ syntax. 2) to overload an
operator, at least one operand has to be a
class argument, see page 283 of the Mar86 printing.
3) It's likely that writing "x->y" would NOT be
interpreted by a human reader as "x to the y power"

Hopefully, if my release of the compiler HAD the
"->" overloading capability, I would have TRIED IT
before I show my mouth off !!!

Given C++ overloading restrictions, the following overload
of "^" shows what CAN be done in C++.  Unfortunately
"^" has poor precedence to be used as a "pow" operator,
so you'd have to parenthesize it:

mycubic = a[3]*(x^3) + a[2]*(x^2) + a[1]*x + a[0];

which MIGHT be better than writing:

mycubic = a[3]*pow(x,3) + a[2]*pow(x,2) + a[1]*x + a[0];

[to use a weak example!]

By analogy to using "<<" as the output operator, using "^"
to mean "pow" MIGHT be justifiable if you need it enough,
and use it enough in your code to make its use obvious
from context.

Again, IFF the power operator is needed by most users of a
language, then it should be in the language, otherwise
leave it out.

P.S. I HAVE tried the following, which is based on the half-baked
    idea I had in the back of my head when I shot my mouth off 
    last time!!!

*/

#include <stdio.h>
extern double pow(double x, double y);

class Double
{
public:
    double dval;
    Double(double val) {dval = val;};
};  

inline double operator^(Double& x, int y) 
{ return pow(x.dval, double(y)); }

main()
{ 
    int i;
    Double x = 2.0;
    
    for (i = 0; i <= 10; ++i) printf("%g \n",x^i);
}