[Potential Feature] `assignAll` notify the listener twice

[alexandruantonica]

Is there any way to make assignAll to do not notify the listeners twice (once at clear and another time at addAll) ?

Is your feature request related to a problem? Please describe. Every time I am trying to listen a RxList I have the problem that I receive 2 events every time I want to "replace" the list with a new one.

final RxList<String> myAwesomeList = RxList();

// onInit
myAwesomeList.listen( (newList) => print('new value : ${newValue.length}'));

// somewhere in the app 
myAwesomeList.assignAll(['This', 'is', 'cool'];

// Once I call `assignAll` the `listen` will return this 
// -> "new value : 0" 
// -> "new value : 3"

// if I call it again 
myAwesomeList.assignAll(['nothing'];
// -> "new value : 0" 
// -> "new value : 1"

Describe the solution you'd like I have tried to understand how the assignAll is working and after I have tried multiple solutions, seems like the only way is to change the set value from RxObjectMixin and force it somehow to not sentToStream in some specific situations. But also to be able to control the sentToStream from assignAll (or create replaceAll, probably someone really needs the current assignAll behaviour), because now sentToStream can be accessed from assignAll but it can't change the behaviour of set value.

Any better idea I can try ?

Temporary Workaround

// instead of `.listen` I use `debounce` worker
debounce<List<String>>( myAwesomeList, (newList) => print('debounce : ${newList.length}'));

// it works because the operations `clear` and `addAll` are quick and ignored by `debounce`, so it will return only the last value provided by `addAll`

[alexandruantonica]

@jonataslaw Do you have any idea how we can solve this issue? Or any other approach I can try to avoid the double firing of the listen callback ?