`ref T` should be a type!!

[Rubn]

On Monday, 1 April 2019 at 00:18:40 UTC, Walter Bright wrote:
> On 3/29/2019 12:55 PM, Rubn wrote:
>> You can't have reference variables in D, and I doubt walter 
>> will allow them. It doesn't make much sense to have a type 
>> that allows ref if you can't declare a variable with it.
>
> You can't declare a pointer to a ref in C++, either. It's not a 
> real type.

It's a type with a restriction, you can literally use it 
everywhere else you can use a type. A reference is just a pointer 
with a dress on, having a naked pointer go with a dressed pointer 
doesn't make much sense.

Taking the address of a reference returns a pointer to the object 
the reference points to, not a pointer to the reference. It's 
impossible to get the address of the actual reference (* without 
some hack).

But sure it's not a type because you can't have a pointer to it.

     std::is_same_v<int, int&>;