Re: [boost] Metaprogramming: Using static const or enum?

From: Gabriel Dos Reis (gdr_at_[hidden])
Date: 2003-01-11 15:05:42

"Paul Mensonides" <pmenso57_at_[hidden]> writes:

| ----- Original Message -----
| From: "Gabriel Dos Reis" <gdr_at_[hidden]>
| 
| > |                                                          However, this
| > | raises another interesting problem.  Many enumerations used for this
| purpose
| > | are unnamed:
| > |
| > | template<class T> struct is_ptr {
| > |     enum { value = false };
| > | };
| > |
| > | template<class T> struct is_ptr<T*> {
| > |     enum { value = true };
| > | };
| > |
| > | This has its own problem in the same templated "pass by reference"
| > | situation.  Namely, a template type argument cannot refer to an unnamed
| > | type:
| > |
| > | template<class T> T func(const T& ref) {
| > |     return ref;
| > | }
| > |
| > | int main() {
| > |     func( is_ptr<int*>::value );
| > |         // error:  argument deduction yields unnamed type
| >
| > Probably I'm being dense here, but please could you work out *why* you
| > think the above should fail?
| 
| The type of "is_ptr<int*>::value" is "is_ptr<int*>::unnamed_type" _not_
| "int".

This, I agree with.

| Therefore template argument deduction should result in:
| 
| is_ptr<int*>::unnamed_type func(const is_ptr<int*>::unnamed_type& ref);

agreed.

| ...which makes "T" equal to "is_ptr<int*>::unnamed_type" -- which is
| illegal.

This, I don't get.  What I am missing?

| The problem is not binding the reference to const (as would be the problem
| with static const without a definition), but rather the template argument
| deduction attempting to bind an unnamed type to a template parameter.

Yes, but which text in the Standard forbids that?

Thanks,

-- Gaby