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
>>
>>
