Re: Access types - what's happening here?

["Yannick Duchêne (Hibou57)" <yannick_duchene@yahoo.fr>]

Le Wed, 26 May 2010 00:27:39 +0200, KK6GM <kk6gm@att.net> a écrit:

> 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 := 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 := 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 it :

If you do

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

You get the same thing as if you do

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