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From: Matthew Heaney <matthew_heaney@acm.org>
Subject: Re: Enumeration literal visibility and use type
Date: 1998/05/26
Message-ID: <m3btsk8xoa.fsf@mheaney.ni.net>#1/1
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Distribution: na
References: <6kej65$dnh$1@hermes.seas.smu.edu>
Organization: Network Intensive
Newsgroups: comp.lang.ada
Date: 1998-05-26T00:00:00+00:00
List-Id: <comp.lang.ada>


bralick@seas.smu.edu (William Bralick) writes:

> For some reason that I have yet to fathom, I had convinced myself 
> that "use type" would give immediate visibility of the enumerals in 
> doodah.cool_type.  I am now convinced that it doesn't (though I am
> ready to argue that it _should_).
> 
> Because sharing this information with my other team members will require
> that I eat more than a little crow ;-), I am looking for a pointer to 
> any information (chapter and verse in the rationale, an AI, anything ...) 
> that would explain _why_ Ada95 does not provide immediate visibility to 
> enumeration types that have been the object of a "use type" clause.

No, use type will NOT give you direct visibility to the enumeration literals.

I helps to understand the type model wrt to enumeration types.  In Ada,
an enumeration literal is modeled as a "primitive operation" of the
type.  Given this declaration,

type T is (E1, E2, E3);

it's as if it were declared like this:

type T;
function E1 return T;
function E2 return T;
function E3 return T;

The reason for this is that it allows different enumeration types in the
same scope to have the same values for literals, ie

type T2 is (E2, E3, E4);

O1 : T1 := E2;
O2 : T2 := E2;

The model is that E2 is a parameterless function returning a value of
the type.  Since each type is different, each "function" is different,
and so there is no confusion about what E2 means.  The usual name
resolution rules apply.

The issue is that "use type" only gives you direct visibility to the
predefined "operators" of a type; it does not give you direct visibility
to the predefined "operations" of a type.  Enumeration literals are
operations, not operators, and so aren't made directly visible by "use
type."

Does that explain it?

Matt

P.S.  Here's another example to illustrate how enumeration types work.
Given this declaration, 

generic
   type T is private; 
   function F return T;
package GP is ...;

I can instantiate it using an enumeration literal, passing it as the
generic actual "function":

package P is new GP (T, E2);

Pretty cool, eh?