http://quantlib.414.s1.nabble.com/Questions-about-the-implementation-of-Observable-Observable-operator-const-Observable-o-tp7820p7822.html
> Q1> I suppose because this is a base class not meant to be used directly,
> but only through derived classes. The default copy constructor _of the
> derived class_ first copies the base object, thus calling
> Observable::operator=, then copies the derived object’s data members.
>
> Q2> The base object isn’t modified (because the class has no actual content
> that should be updated), but the derived object is.
>
>
>
> Gerardo
>
>
>
>
>
> Da: Mike Jake [mailto:
[hidden email]]
> Inviato: mercoledì 30 maggio 2012 18.02
> A:
[hidden email]
> Oggetto: [Quantlib-users] Questions about the implementation of Observable&
> Observable::operator=(const Observable& o)
>
>
>
> /*! \warning notification is sent before the copy constructor has
> a chance of actually change the data
> members. Therefore, observers whose update() method
> tries to use their observables will not see the
> updated values. It is suggested that the update()
> method just raise a flag in order to trigger
> a later recalculation.
> */
> inline Observable& Observable::operator=(const Observable& o) {
> // as above, the observer set is not copied. Moreover,
> // observers of this object must be notified of the change
> if (&o != this)
> notifyObservers();
> return *this;
> }
>
>
> Question 1> Why the doc says that "notification is sent before the copy
> constructor has a chance of actually change the data members"?
>
> Question 2> Since the operator= doesn't modify the current object(this), why
> we should call notifyObservers()?
>
> Thank you
>
>
>
>
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