Invariant doesn't apply to declared symbols

[Janice Caron]

> So here's my latest and greatest idea. What about:
>
>     const(C)ref // just the data is const
>     const(C) // the data and the pointer are both const
>
> This would give us
>
>     const(C) c;
>     const(C) ref d;
>
>     c = new C: // Error;
>     d = new C; // OK
>     d.x = 100; // Error

Or...

Since we like function-style type constructors in D:

     ref(const(C)) // just the data is const
     const(C) // the data and the pointer are both const

In the interests of plain old fashioned common sense, the following
four lines should all be exactly equivalent and interchangeable...


    ref const C c;
    ref const(C) c;
    ref(const C) c;
    ref(const(C)) c;

In each case, we are declaring c to be of type C, and then saying that
c is const - except for the actual reference, which isn't. Note that
ref-as-a-type-constructor would naturally be forbidden for
non-reference types, so

    ref(const(int)) n;

would be a syntax error, because int isn't a reference type.

(...at least, until we get references in D, at which point we will
then be able to legalise it).