----- Original Message -----
From: "David Abrahams" <[EMAIL PROTECTED]>
> > solution:
> >
> > There is no non-intrusive solution. It must be cast at the call site to
a
> > common type (such as 'int' or 'bool') in order to avoid replicated
template
> > instantiation.
> >
> > ---- */
>
> Of course, you can use the first solution with enums as well:
>
> template <class T, T N>
> struct integral_c
> {
> enum value_type { value = N; };
> };
>
> template<class T> struct Y
> : integral_c<int,10> {};
>
>
> Now Y<int>::value and Y<double>::value have the same type. Of course,
> you will probably want some additional tools in integral_c<T,N>, like
> an explicit conversion to an rvalue of type T.
Well, yes. If you do that, however, it removes one of the primary reasons
to use enumerations, which is for the syntactic convenience. The only
reason left to use enums is to avoid the static storage of the values. The
automatic rvalue conversion would be nice, but you'd still get the template
instantiation problem if you did this:
template<class T> void f(const T&);
template<class T> class is_integer : integral_c<bool, false> { };
template<> class is_integer<int> : integral_c<bool, true> { };
template<class T> void f(const T&);
int main() {
f( is_integer<int>() ); // 'T' == 'is_integer<int>'
return 0;
}
What you need is both 'value' and 'rvalue':
template<class T, T V> struct integral_c {
enum value_type { value = V };
static inline T rvalue() {
return value;
}
};
int main() {
f( is_integer<int>::rvalue() ); // better
return 0;
}
This still doesn't solve one other issue though:
template<class T> T operator && (T, T); // etc.
...which is evil, I agree, but possible. This can still break any
arithmetic or logical operations that involve enumerations:
is_integer<int>::value && is_integer<int>::value // not constant
The only solution that I see is:
(bool)is_integer<int>::value && (bool)is_integer<int>::value
Paul Mensonides
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