Implicit dereferencing

[Steven Schveighoffer]

On Mon, 01 Apr 2013 23:23:49 -0400, Luís Marques <luismarques at gmail.com>  
wrote:

> On Tuesday, 2 April 2013 at 02:52:48 UTC, Steven Schveighoffer wrote:
>> You see, indexing does NOT dereference the pointer, it's an index for  
>> that pointer.  c[0] means *(c + 0).  A pointer is essentially an  
>> unchecked slice, with undefined length.  This is how it works in C also.
>>
>> c[1] is the same as *(c + 1), completely consistent (and also sets b to  
>> 42, 42)
>
> OK, I think I see where I went astray. I was a case of bad induction  
> from a few tests :-)
> So, I guess what is happening is the following, right?
>
>      int[2] a;
>      int[2] *c;
>      c = &a;
>
>      c[0] = 7;  // same thing as below
>      a = 7;      // same thing above
>
>      (cast(int*) c)[0] = 7; // but different from this
>
> I verified that c is a pointer to a.ptr, I guess what I didn't consider  
> is that because c points to int[2], the assignment becomes the same as a  
> = 7, and not a[0] = 7.
>
> Still, what do you think of the struct vs AA automatic pointer  
> dereferencing?

a pointer defines indexing.  To have implicit dereferencing for indexing  
on any pointer type would not be good.

For example:

string[int][2] aas;
string[int] *aaptr = aas.ptr;

auto x = aaptr[1];

If x is not a pointer to aas[1], then you would have to access it using  
*(aaptr + 1), which would suck.  Plus it would make pointers to indexed  
types inconsistent with all pointers to types that don't define indexing.

-Steve