Why no acess to other structs in classes?

[Bill Baxter]

BCS wrote:
> Accessing the outer class would require a pointer to it. The only way to get that
> pointer would be to add a hidden value to the struct. In D, structs are bare bones
> aggregates without any hidden stuff added in.
> 
> == Quote from Karen Lanrap (karen at digitaldaemon.com)'s article
>> class C
>> {
>>     struct S
>>     {
>>         uint data;
>>     }
>>     S s;
>>     struct T
>>     {
>>         uint f()
>>         {
>>             return s.data;
>>             // this for s needs to be type C not type T *
>>         }
>>     }
>> }
>> void main(){}
> 

I don't think so.  Structs are always treated like plain old data.  So 
in memory this:
    class C
    {
       int a;
       int b;
    }
has exactly the same layout as
    class C
    {
       int a;
       struct S { int b; }
       S s;
    }

Therefore, if it is possible for S to compute the address to it's own 
members:
    class C
    {
       int a;
       struct S {
          int b;
          int f() { return b+1; }
       }
       S s;
    }

Then it should also be possible for S to compute the address of members 
in the outer class at compile time.

Try running this program:
------------
import std.stdio : writefln;

class C
{
     int a;
     struct S {
         int b;
         int outer_a() {
            return cast(int)* (cast(char*)(&b)-int.sizeof);
         }
     }
     S s;
}

void main()
{
     C c = new C;
     c.a = 42;
     writefln("c.a's value is ", c.s.outer_a());
}
---------------

The offset to the members of the outer class (or outer struct) are 
compile time constants.  It's well within the compiler's ability to 
compute them.

--bb