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From: Gennaro Prota <gennaro_prota@yahoo.com>
Newsgroups: comp.std.c++
Subject: Re: Why switch w/ only integral types?
Date: Wed,  1 May 2002 17:15:30 GMT
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On Tue, 30 Apr 2002 18:44:01 GMT, David Schwartz
<davids@webmaster.com> wrote:

[...]
>
>	There are reasons. Consider:
>
>MyFoo f;
>
>switch (f)
>{
> case 1:
>  // code
> case 2:
>  // code
>}
>
>	What should happen if MyFoo::operator==(int) returns 'true' for both 1
>and 2?
>

I think you completely missed the point.
We are talking about using non-integral expressions in case labes, not
in the switch condition.

With the current rules, the following is legal:


class A {
    public:
	typedef char whatever;
	A(whatever) {}
	operator int() const { return 1;}
};
	
	
switch (A a = 'x') {
	case 1:
		/*...*/
	case 2:
		/*...*/
	default:
}

and causes A::operator int() const to be invoked at the evaluation of
A a = 'x' . No operator ==() is currently needed.


P.S.: for Greg Comeau (I hope you are reading...). Comeau online seems
to accept the above switch statement in some cases where A has more
than one conversion function to integral/enumeration types, for
instance:

       operator int() const { return 1;}
       operator char() const {return 'a';}
       operator bool() {return true;}


Anyhow it emits an error if I declare also operator bool() as a
const-function. Isn't this a bug?


Genny.

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