What's the least horrible way to remove all listen...

# tornadofx

[Ahmed Mourad]

What's the least horrible way to remove all listeners on a property? without having to keep reference to each and every one of them.

[Marshall]

The easiest way is to just dispose of the object 😛

[Marshall]

otherwise, you either have to keep a list of all the listeners, use bindings instead, or use a different strategy

[Ahmed Mourad]

Unfortunately, I need a property not a binding, I guess I could just dispose the object 😐 Tried all weird things with weak references but didn't think of that for some reason 😅

[Marshall]

of course you can bind the property to other properties or a function/lambda but that may not work depending on what you are doing

[Ahmed Mourad]

Yup, it's a bit complicated, I was trying to bind two properties of different types bidirectionally, one of which will be bound bidirectionally to other properties of other types very frequently.

To achieve it I tried adding listeners to both properties and propagating the values between them, but it was creating memory leaks even when I used `WeakChangeListener`s.

[Marshall]

you can use "bindBidirectional" and then unbind

[Marshall]

val test = booleanProperty(false)
val other = booleanProperty(false)
test.bindBidirectional(other)

[Marshall]

you just have to call unbind() on both or test.unbindBidirectional(other)

[Ahmed Mourad]

Yup, but I can't pass a converter for bidirectional binding, as far as I'm aware, so it works only with properties of the same type.