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

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

On 5/5/20 1:48 PM, Niklas Holsti wrote:
> On 2020-05-05 14:04, hreba wrote:
>> 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;
>>
>> Compiling results in "non-local pointer cannot point to local object" 
>> indicating the marked line above.
>>
>> How can one record component be local and another not?
>>
>> Intention is to have a library subprogram Init() for initialization. 
>> How would I implement that instead?
>>
>> (The reason for the above construction is the need to pass a pointer 
>> to a C library function.)
>>
> 
> The parameter "r" might be passed by copy, in which case the r.a'Access 
> would point to the local copy, and become a dangling pointer on return 
> (copy-out).

Now I understand.

> It works if you replace "r: in out" with "r : access", which ensures 
> pass by reference.

Perfect, works indeed. Thanks a lot.

I tried another thing too: Changed Init into a constructor function 
returning r in an extended return statement. Didn't work. So even an 
extended return statement seems to create a local copy first.

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