Re: Question on controlled types

["Alex Xela" <xela2@free.fr>]

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Thanks, it is what I was expecting (where did you find the answer, in the
RM95?).

The example below is trying to synthesize your answer:



--------------------------------------------------
with Ada.Finalization;
package Instrum is
    type Instrum_Type is new Ada.Finalization.Controlled with
    record
        A : Integer;
    end record;
    procedure Finalize  (O : in out Instrum_Type);
    procedure Adjust  (O : in out Instrum_Type);
    procedure Initialize  (O : in out Instrum_Type);
    J : Instrum.Instrum_Type := ( Ada.Finalization.Controlled with
A=>0);--nothing : OK
    --J : Instrum.Instrum_Type; --raised PROGRAM_ERROR : access before
elaboration : OK
end Instrum;
--------------------------------------------------
with Instrum, Ada.Finalization,Ada.Text_Io;
procedure Test_Instrum is
    I : Instrum.Instrum_Type := Instrum.J; --adjust : OK
    --I : Instrum.Instrum_Type := ( ada.finalization.controlled with
A=>0); --nothing : OK
    --I : Instrum.Instrum_Type; --initialize : OK
begin
    Ada.Text_Io.Put_line("end;");
end Test_Instrum;


"Matthew Heaney" <matthewjheaney@earthlink.net> a �crit dans le message de
news: u8yo4imig.fsf@earthlink.net...
> "Alex Xela" <xela2@free.fr> writes:
>
> > On the following code, what should happen during J variable elaboration?
> >
> >  - Nothing
> >
> >  - A call to Adjust.
> >
> > I tried to find the answer in RM95 but in vain.
> >
> > My tests on several compilers (and my feeling) were suggesting  :
nothing!
>
> Aggregate assignment is handled specially.  For a controlled type, it
> means the object must be built in place.
>
> Clearly, no controlled operations can be called, because they haven't
> been elaboratated yet.  Hence the special rule for aggregate assignment.
>
>
>
>