Why can't we make reference variables?

[Tommi]

On Wednesday, 29 August 2012 at 03:21:04 UTC, Jonathan M Davis 
wrote:
>> >> > void main()
>> >> > {
>> >> > 
>> >> >     immutable(Test)* ptr = new immutable(Test);
>> >> >     ptr.foo();
>> >> > 
>> >> > }
>> >> 
>> >> Now, that's a surprise for someone coming from C++. But even
>> >> though ptr looks like a reference variable in your example, 
>> >> it
>> > 
>> >> doesn't look like it at all in this example:
>> > I've been primarily a D guy for years, and even I'm surprised
>> > by that!
>> > O_O
>> 
>> You didn't know that the dot operator does dereference? That's
>> quite a big one to miss for years.
>
> Yeah. I'm a bit confused about what's so suprising about that 
> code.
>
> - Jonathan M Davis

The weird thing is that you can use a member access operator with 
a pointer (without explicitly dereferencing the pointer first). 
At least I didn't know what to expect the following code to print:

struct MyStruct
{
     int _value = 0;

     void increment()
     {
         ++_value;
     }
}

void increment(ref MyStruct* ptr)
{
     ++ptr;
}

void main()
{
     MyStruct[2] twoStructs;
     twoStructs[1]._value = 42;

     MyStruct* ptrFirstStruct = &twoStructs[0];

     // Are we incrementing the pointer using UFCS or
     // are we calling the member function in MyStruct?
     ptrFirstStruct.increment();

     // This prints 1, so we called the actual method
     writeln((*ptrFirstStruct)._value);
}