Thanks for your answer!

One thing, I think, we can agree on, is that it is overly complex for an OO 
language to decorate things like these listeners.
What about introducing another interface like ChangeListenerDecoration with a 
special named „boolean decorates(ChangeListener)) method and call this if the 
listeners hold by *ExpressionHelp class implements that interface?

BTW: If we care about symmetry, why not change the 
listener.equals(otherListener) to listener == otherListener at all? Then, there 
are not implementation details one can make use of.
I even might argue that it is a security risk to use the .equals() the way it 
is now, as an evil listener implementation is able to remove all listeners from 
the list by simply returning true from equals()
It should be the responsibility of the listener in the list to know if the 
passed in listener justify the removal of itself, right? … from a security 
standpoint! :-) 

Regards,
Mario


Am 24.03.2014 um 07:31 schrieb Martin Sladecek <martin.slade...@oracle.com>:

> Mario,
> thanks for your suggestion, but I think your specific case will not justify 
> the change. First of all, as it was already said, equals should be symmetric. 
> Knowing the implementation, you could rely on the fact that either the equals 
> is called on the provided listener or on the listeners already in the list. 
> Might be that somebody else would need the asymmetry to be implemented in the 
> "listener" provided to removeListener() method and the current situation 
> suits him, so why do the change for your case?
> 
> Anyway, relying on some implementation-specific behaviour is generally a sign 
> of bad code practice, so it should give you a hint that you should find a 
> different solution (like the one you discussed with Tomas already). It might 
> happen that the code would be changed / re-factored in the future or some bug 
> would be fixed there and behaviour will change again, breaking your 
> application in some future JFX version.
> 
> Regards,
> -Martin
> 
> 
> On 22.3.2014 15:47, Mario Ivankovits 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<mailto: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><mailto: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><mailto: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><mailto: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
>> 
>> 
>> 
>> 
>> 
>> 
>> 
> 

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