`ref T` should be a type!!

[Rubn]

On Monday, 1 April 2019 at 13:57:17 UTC, Atila Neves wrote:
> On Monday, 1 April 2019 at 13:09:28 UTC, Dein wrote:
>> On Monday, 1 April 2019 at 01:18:44 UTC, Walter Bright wrote:
>>> On 3/31/2019 5:35 PM, Rubn wrote:
>>>> you can literally use it everywhere else you can use a type.
>>>
>>> No, you can't. An array of refs won't compile, either.
>>>
>>>   void test(int& a[]); // error
>>>
>>> A C++ ref can only appear at the top of a type AST, which is 
>>> unlike any other type. Which exactly matches the only place a 
>>> storage class can be!
>>
>> Seriously? You should know as well as I do what that function 
>> actually translates to:
>>
>>    void test(int&* const a);
>>
>> So I ask you now, how do you get a pointer to a reference. Its 
>> the exact same thing. Not sure if you are trying to just 
>> deceive me with some syntax sugar in C++ or what.
>
> Bad example, but the point stands:
>
>
> ----
> // foo.cpp
> int fun() {
>     int& foo[5]; // doesn't compile
> }
> ----
>
> I'd never even thought of it until Walter mentioned it that one 
> can't have an array of references. Huh.

Arrays in C++ are just pointers. If you can't have a pointer to 
something, it is only natural you can't have array either.

     void* ptr;   // ok
     void arr[5]; // error

So I guess Walter agrees that void isn't a type then right? You 
can't have an array to it so that's the deciding factor.