`ref T` should be a type!!

[Rubn]

On Monday, 1 April 2019 at 20:51:44 UTC, Walter Bright wrote:
> On 4/1/2019 1:00 PM, Rubn wrote:
>> Arrays in C++ are just pointers.
>
> No, they're not.
>
>
>> If you can't have a pointer to something, it is only natural 
>> you can't have array either.
>
> g++ -c test.cpp
> test.cpp: In function void test():
> test.cpp:1:24: error: declaration of p as array of references
>  void test() {  int& p[3]; }
>                         ^
>
> As I said, C++ ref can only appear at the top of an AST, which 
> is what a "storage class" is.
>
> C++ tries to have it be both a floor wax and a desert topping, 
> making for a device that nobody understands. Be careful what 
> you wish for.

Let's put it this way, is const a type? No? You would say it is a 
"sturage class". Considering the following now though:

What do you think the following code is going to print?

     import std.stdio;

     void foo(T)(T a) {
         writeln(T.stringof);
     }

     template is_same(T, G) {
         enum is_same = is(T == G);
     }

     void main() {
         int a;
         const int b;

         foo( a );
         foo( b );

         writeln( "is_same: ", is_same!(typeof(a), typeof(b)) );
     }