Re: Why is the destructor called multiple times after I declare an object?

["Dmitry A. Kazakov" <mailbox@dmitry-kazakov.de>]

On 12/01/2016 00:44, Randy Brukardt wrote:
> "Dmitry A. Kazakov" <mailbox@dmitry-kazakov.de> wrote in message
> news:n715oq$fae$1@gioia.aioe.org...
>> On 2016-01-11 19:17, Bob Duff wrote:
>>> Brian Drummond <brian@shapes.demon.co.uk> writes:
>>>
>>>> Somehow I expected "extended return" to allocate space and "build in
>>>> place" during the execution of the return statement.
>>>
>>> Build-in-place is done for return of immutably-limited types,
>>> whether or not the extended return syntax is used.
>>
>> But you can leave one return statement on an exception, catch the
>> exception, and then return through another return statement, with other
>> discriminants and even other type of the result (if it is class-wide).
>>
>> Therefore the result can be potentially allocated and reallocated any
>> number of times. In which sense is this behavior 'in-place'?
>
> "Build-in-place" means simply that there is no temporary object, the object
> is created directly in its final resting place (that is the memory where it
> will live during its existence).
>
> Whatever amount of effort it takes to figure out that final resting place is
> not part of the equation.

Yes, but you said 'no temporary object' which is untrue because a return 
statement creates exactly such an object. You could even use this object 
in order to create another one:

    return Candidate_1 : T (Size => 1) do
       Size := Candidate_1.Get_Required_Size;
       raise Try_Again;
    end return;
exception
    when Try_Again =>
       return Candidate_2 : T (Size => Size) do
           ...
       end return;

Since Ada lacks proper user-defined constructors one cannot claim that 
Candidate_1 was not fully constructed. Even its Initialize was through.

-- 
Regards,
Dmitry A. Kazakov
http://www.dmitry-kazakov.de