Why no offsetof for static struct?

[Mike Parker via Digitalmars-d-learn]

On 7/11/2017 6:14 AM, FoxyBrown wrote:
> On Monday, 10 July 2017 at 20:13:46 UTC, Adam D. Ruppe wrote:

>> No, it isn't. Static members are stored in an entirely different place 
>> than non-static members. They are really just global variables in 
>> memory with their in-source name being nested somewhere else.
> 
> 
> This is not true, just because there isn't any official statement, or 
> even if denied, does not make it true. I have proof:
> 
> auto GetStaticAddress(T)()
> {
>      mixin("auto p = cast(T*)&T."~__traits(allMembers, T)[0]~";");
>      return p;
> }
> 
> 
> Returns the address of a struct's static members.
> 
> It's pretty obvious, the compiler seems to instantiate the static 
> members simply as a sort of "singleton". They are laid out with the same 
> relative offsets as the struct(which is obvious, as that is the most 
> natural way). That is all that is needed to treat the memory the static 
> variables use as the struct in which they are part of. No need to make 
> it any more complex than that.
> 

Every variable has an address. That includes static members of a struct. 
But their location has absolutely no relationship to any instance of a 
struct. They have a fixed location in memory. Consider:

module Foo;
int x;
struct Bar {
     static int y;
     int z;
}

There is effectively no difference between x and y here. The only 
difference is that Bar is part of y's namespace, so you can access them 
like so:

Foo.x;
Foo.Bar.y;

y's location has no relationship to any instance of Bar, and the 
declaration of the Bar type has no address. Of course you can take the 
address of y, as &Bar.y, but you can also take the address of x. Take y 
out of Bar and the only difference is that you no longer need Bar as 
part of the namespace.  On the other hand:

Bar b;
writeln(b.z);

As an instance of Bar, b *does* have a location in memory. And z *does* 
have an offset that is relative to that location.

So y.offsetof does not exist, because there is no instance of Bar to 
which it can have a relative address.