From mboxrd@z Thu Jan  1 00:00:00 1970
X-Spam-Checker-Version: SpamAssassin 3.4.4 (2020-01-24) on polar.synack.me
X-Spam-Level: 
X-Spam-Status: No, score=-0.8 required=5.0 tests=BAYES_00,INVALID_DATE
	autolearn=no autolearn_force=no version=3.4.4
X-Google-Language: ENGLISH,ASCII-7-bit
X-Google-Thread: 103376,9697af769c7c74cc,start
X-Google-Attributes: gid103376,public
X-Google-ArrivalTime: 1993-04-01 17:39:28 PST
Newsgroups: comp.lang.ada
Path: 
 gmd.de!ira.uka.de!sol.ctr.columbia.edu!zaphod.mps.ohio-state.edu!darwin.sura.net!udel!news.intercon.com!psinntp!scylla!davidg
From: davidg@oracorp.com (David Guaspari)
Subject: Language lawyer needed
Message-ID: <1993Apr1.215753.3147@oracorp.com>
Sender: davidg@oracorp.com (David Guaspari)
Organization: ORA Corporation, Ithaca, New York
Distribution: usa
Date: Thu, 1 Apr 1993 21:57:53 GMT
Date: 1993-04-01T21:57:53+00:00
List-Id: <comp.lang.ada>

Here's a question for language lawyers.

Consider the following pathological program:

   procedure Q(x : integer) is  
   begin
     declare
       x : integer;   -- (1)
     begin
       Q(x => 0);     -- (2)
     end;
   end Q;

My question concerns the legality of Q (forgetting about the fact that
its execution won't terminate).  One way to apply chapter 8 in the
reference manual goes as follows:

  The declaration of x in (1) hides the declaration of x as a formal
  parameter of Q.

  Therefore, by the visibility rules, the occurrence of x in (2) has
  exactly one possible meaning, namely that given by the declaration
  in (1). 

  However, the declaration in (1) is not visible by selection at the
  point at which x occurs in (2).

  Therefore the call in (2) is illegal.

One the other hand, at least one compiler (Telesoft) thinks that Q is
legal, an outcome that seems a reasonable enough outcome to common
sense untutored by the reference manual.  So, what *is* the right
answer, and how do we get it from the RM?

- David Guaspari
  davidg@oracorp.com