JavaFX:绑定(bind)和弱监听器

Question

来自 Javadoc for bind() :

Note that JavaFX has all the bind calls implemented through weak listeners. This means the bound property can be garbage collected and stopped from being updated.

现在考虑我有两个属性 ObjectProperty<Foo> shortLived居住在ShortLivedObject和 ObjectProperty<Foo> longLived居住在LongLivedObject .

我像这样绑定(bind)它们:

longLivedObject.longLivedProperty().bind(shortLivedObject.shortLivedProperty());

因为绑定(bind)使用弱监听器，所以如果shortLivedObject是否收集了垃圾，shortLived属性(property)也将被垃圾收集。那么，这是否意味着 longLived属性仍然绑定(bind)，但永远不会更新？那会留下吗longLived处于绑定(bind)状态的属性(使进一步绑定(bind)变得不可能)，但什么都不做？

Answer 1

So, does that means that longLived property is still bound, but it will never be updated?

假设 shortLivedProperty 已被垃圾回收，shortLivedProperty 将永远不会再次失效。因此，longLived 的监听器将永远不会被再次调用和更新。

Does that leave longLived property in a bound state (making further binding impossible), but does nothing?

无论绑定(bind)状态如何，您都应该始终能够将属性绑定(bind)到新 observable，因为旧的 observable 属性将为您删除/解除绑定(bind):

public void bind(final ObservableValue<? extends T> newObservable) {
    if (newObservable == null) {
        throw new NullPointerException("Cannot bind to null");
    }

    if (!newObservable.equals(this.observable)) {
        unbind();
        observable = newObservable;
        if (listener == null) {
            listener = new Listener(this);
        }
        observable.addListener(listener);
        markInvalid();
    }
}