Hi Mario,
even if your proposal gets accepted, you would still depend on an
implementation detail of JavaFX, which is always good to avoid.
To sum up, my second proposal compared to your current solution:
(+) your equals() stays symmetric;
(+) no need for an extra Map of listeners.
My second proposal compared to your desired solution if your proposal
is accepted:
(+) your equals() stays symmetric;
(+) you don't depend on an implementation detail in JavaFX.
Best,
Tomas
On Sat, Mar 22, 2014 at 3:47 PM, Mario Ivankovits <ma...@datenwort.at> wrote:
> The only thing which I ask for is to flip this „if" in the *ExpressionHelper
> classes:
> So, JavaFX does not break anything and it is up to the app developer to take
> the risk or not.
>
> if (listener.equals(changeListeners[index])) {
>
> If we flip this to
>
> if (changeListeners[index].equals(listener))
>
>
>
>
> Am 22.03.2014 um 15:42 schrieb Kevin Rushforth <kevin.rushfo...@oracle.com>:
>
> If you are talking about a change to the JavaFX API, we are not likely to
> accept such a change if it breaks the contract of equals() or hashcode(). In
> your own app it may not matter, but it is dangerous to assume it won't
> matter to anyone. Martin owns the core libraries and can comment further.
>
> -- Kevin
>
>
> Mario Ivankovits wrote:
>
> Hi Thomas!
>
> Thanks for your input. Because I want to decorated listeners added by JavaFX
> core I can not use the sub/unsub pattern.
> Your second proposal is almost what I do right now. In removeListener I
> consult a map where the decorated listeners and their undecorated one lives.
>
> Regarding the symmetric doubts. Such listeners will always be removed by
> passing in the undecorated object to removeListener.
> They should act like a transparent proxy.
>
> Even if this breaks the equals paradigm, in this special case I can
> perfectly live with an equals/hashCode implementation like this below.
> It won’t break your app; as long as both objects do not live in the same
> HashSet/Map …. for sure - but why should they?
>
> @Override
> public boolean equals(Object obj)
> {
> return obj == delegate; // && this == obj
> }
>
> @Override
> public int hashCode()
> {
> return delegate.hashCode();
> }
>
>
> Regards,
> Mario
>
>
> Am 22.03.2014 um 14:22 schrieb Tomas Mikula <tomas.mik...@gmail.com>:
>
>
>
> A simpler and more universal solution is to also override removeListener:
>
> public void removeListener(ChangeListener<? super T> listener) {
> super.removeListener(decorated(listener));
> }
>
> And the equals() method on decorated listeners only compares the
> delegates (thus is symmetric).
>
> Regards,
> Tomas
>
>
> On Sat, Mar 22, 2014 at 2:07 PM, Tomas Mikula <tomas.mik...@gmail.com>
> wrote:
>
>
> The suspicious thing about your solution is that your smart
> implementation of equals() is not symmetric.
>
> In case the observable value is visible only within your project, you
> could do this:
>
> interface Subscription {
> void unsubscribe();
> }
>
> class MyObservableValue<T> implements ObservableValue<T> {
> public Subscription subscribe(ChangeListener<? extends T> listener) {
> ChangeListener<T> decorated = decorate(listener);
> addListener(decorated);
> return () -> removeListener(decorated);
> }
> }
>
> and use subscribe()/unsubscribe() instead of addListener()/removeListener():
>
> Subscription sub = observableValue.subscribe(listener);
> // ...
> sub.unsubscribe();
>
> Of course this is not possible if you need to pass the observable
> value to the outside world as ObservableValue.
>
> Regards,
> Tomas
>
> On Sat, Mar 22, 2014 at 6:57 AM, Mario Ivankovits <ma...@datenwort.at>
> wrote:
>
>
> Hi!
>
> In one of my ObservableValue implementations I do have the need to decorate
> ChangeListener added to it.
> Today this is somewhat complicated to implement, as I have to keep a map of
> the original listener to the decorated one to being able to handle the
> removal process of a listener. Because the outside World did not know that I
> decorated the listener and the instance passed in to removeListener ist the
> undecorated one.
>
> We can make things easier with a small change in
> ExpressionHelper.removeListener. When iterating through the listener list,
> the listener passed in as argument to removeListener is asked if it is equal
> to the one stored
>
> if (listener.equals(changeListeners[index])) {
>
> If we flip this to
>
> if (changeListeners[index].equals(listener)) {
>
> a listener decoration can be smart enough to not only check against itself,
> but also against its delegate.
>
> What do you think?
>
> I could prepare a patch for the *ExpressionHelper classes.
>
>
> Best regards,
> Mario
>
>
>
>
>
>
>
