Re: Initialize with aggregate?

["Adam Beneschan" <adam@irvine.com>]

Robert A Duff wrote:
> ejijott@gmail.com writes:
>
> > Hi there!
> >
> > Regarding the code below, what syntax do I use to initialize tempstore
> > with an aggregate?
> > [code]
> > .
> > .
> > .
> >     type node;
> >     type storage is access node;
> >     type node is
> >         record
> >             next:  storage;       -- Pekare till nästa nod.
> >             item:  string(1..50); -- En textsträng
> >             len:   integer :=-1;  -- Längden på strängen.
> >             count: integer := 0;  -- Antalet förekomster av strängen
> >         end record;
> > .
> > .
> > .
> >         tempstore:=new Node;
> >         tempstore.item(1 .. Val'length):=Val;
> >         tempstore.len:=Val'length;
> >         tempstore.count:=1;
> >         if S /= null then
> >             tempstore.next:=S;
> >         end if;
> >         S:=tempstore;
> > [/code]
>
> How about something like this:
>
>     type node;
>     type storage is access node;
>     type node(len: natural) is
>         record
>             next:  storage;       -- Pekare till nästa nod.
>             item:  string(1..len); -- En textsträng
>             count: integer := 0;  -- Antalet förekomster av strängen
>         end record;
>
>     tempstore:=new Node'(
>             len => val'length,
>             next => S,
>             item => val,
>             count => 1);
>     S:=tempstore;
>
> Hardwiring Item'Length to 50 isn't a good idea, and it makes the
> programming harder, because you have to carefully avoid the junk
> at the end.

I don't see how you can make a blanket statement like this.  If you
code it the way you've shown, you can't change "len" in a node once
you've allocated it.  But sometimes changing "item" and "len" in an
existing node to change the length of the string, without changing the
other fields in the node, is what you want to do, so there are cases
where the original poster's implementation is more appropriate than
using a discriminant record.  (Using Ada.Strings.Bounded may make
programming easier still, but you still have to hardwire a maximum
length.)

                                    -- Adam