Re: Language lawyer question: Equality on 'Access attributes

[Robert A Duff <bobduff@shell01.TheWorld.com>]

"Robert I. Eachus" <rieachus@comcast.net> writes:

> So let me take the example and modify it a bit:
> 
>   package Pak1 is
>       type Ptr1 is access Integer;
>       X : aliased Integer;
>       Y : aliased Integer;
>       B : Boolean := X'Access = Y'Access;
>   end Pak1;

I guess all you did was erase the "all"?

> Is your position Bob, that this version is illegal?

Yes.  There's only one "=" of interest here.  So the expected type of
X'Access is Ptr1, which is not a general access type.  You can't do
X'Access returning a non-general access type.

In other words, overload resolution works the same as in the other
example, but there's a legality rule being violated.

>...Now there is no
> visible general access type, or at least none that the user can see.  

Well, I'm presuming you're showing me the whole example!  If there are
some other "=" operators visible that I can't see, then who knows?

> What is the expected type of "=" above?

Boolean.  Not sure what you mean...

>...Can we use the fact that
> anonymous access types don't have equality operators.

I don't see any anonymous access types in the example.
But yes, if there were, the "=" operators that are visible
would not include ones for those anonymous types.
And resolution would proceed accordingly.

> Whew!  First, 3.10.2(2) was a Name Resolution Rule, so it can't make a
> construct illegal.  However, 4.5.2(6) is "Static Semantics."  Can it be
> considered in a name resolution context?

Yes.

>...  And if so does it make a
> construct illegal, or rule it out of consideration.

It just tells us which implicit declarations exist.
That information affects overload resolution.

- Bob