Re: How can one record component be local and another not?

[hreba <f_hreba@yahoo.com.br>]

On 5/5/20 1:33 PM, AdaMagica wrote:
> Am Dienstag, 5. Mai 2020 13:04:56 UTC+2 schrieb hreba:
>> I have reduced my problem to the following example:
>>
>> package Aux is
>>      type Integer_P is access all Integer;
>>      type Rec is record
>>         a: aliased Integer;
>>         p: Integer_P;
>>      end record;
>>
>>      procedure Init (r: in out Rec);
>> end Aux;
>>
>> package body Aux is
>>      procedure Init (r: in out Rec) is
>>      begin
>>         r.p:= r.a'Access;  -- <-- error!
>>      end Init;
>> end Aux;
> 
> Imagine you define a local Rec:
> 
> with Aux;
> procedure Proc is
>    R: Aux.Rec;
> begin
>    Aux.Init (R);
> end Proc;
> 
> Now R has shorter lifetime than the access type Integer_P. Thus you might have somewhere an access object AO with longer lifetime than R and try to assign
> AO := R.P;
> 
> Try Unrestricted_Access.
> 

Don't understand that. R.a and R.p have exactly the same lifetime, which 
is the lifetime of R, haven't they?

-- 
Frank Hrebabetzky, Kronach	+49 / 9261 / 950 0565