Re: Question on Controlled types

["Hibou57 (Yannick Duchêne)" <yannick_duchene@yahoo.fr>]

On 30 mai, 15:27, "Dmitry A. Kazakov" <mail...@dmitry-kazakov.de>
wrote:
> The formula is I + A = F, where I is the number of calls to Initialize, A
> is the number of calls to Adjust and F is the number of calls to Finalize.
>
> Note that the compiler has right to optimize some pairs Adjust/Finalize
> away or add new ones.

So this means that if there was a second assignation of Delay1,
Finalize would be invoked before the copy and the Adjust which will
follow the copy ?

With one single assignation :

1) Delay1 assigned
2) Adjust invoked

3) Finalize invoked

With two assignations:

1) Delay1 assigned
2) Adjust invoked

3) Finalize invoked before copy
4) Delay1 assigned
5) Adjust invoked after copy

6) Finalize invoked

Is that right ?