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From: =?iso-8859-15?Q?Yannick_Duch=EAne_=28Hibou57=29?=
 <yannick_duchene@yahoo.fr>
Newsgroups: comp.lang.ada
Subject: Re: Access types - what's happening here?
Date: Wed, 26 May 2010 00:38:03 +0200
Organization: Ada At Home
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Date: 2010-05-26T00:38:03+02:00
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Le Wed, 26 May 2010 00:27:39 +0200, KK6GM <kk6gm@att.net> a =E9crit:

> Looking at Cohen's "Ada as a 2nd Language" I see in chapter 8 (Access
> Types), p.319, this construct which I do not understand.  But first,
> the construct I do understand:
>
>   Old_Serial_Number :=3D Serial_Number("Lovelace, Ada").all;
>
> No confusion about that whatsoever.  However, that is followed by
> this:
>
>   "More interestingly, the assignment
>       Serial_Number( "Lovelace, Ada").all :=3D New_Serial_Number;
>   copies the contenst of the integer variable into the variable
> pointed to by the result of the function call."
>
> Huh?  Where does the New_Serial_Number value end up here?

As I don't see any trouble here, I'm not sure I've understood the  =

question. However, here is the answer I give to it as I can understand i=
t :

If you do

    Serial_Number( "Lovelace, Ada").all :=3D New_Serial_Number;

You get the same thing as if you do

    Old_Serial_Number :=3D Serial_Number("Lovelace, Ada").all;
    Old_Serial_Number.all :=3D New_Serial_Number;

Can you give a link to the page you are reading ?

-- =

There is even better than a pragma Assert: a SPARK --# check.