关于java：JavaFX：绑定和弱监听


JavaFX: Binding and weak listener

来自 bind() 的 Javadoc:

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 驻留在 ShortLivedObject 和 ObjectProperty<Foo> longLived 驻留在 LongLivedObject.

我是这样绑定它们的：

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

因为绑定使用弱监听，所以如果 ShortLivedObject 被垃圾回收，shortLived 属性也会被垃圾回收。那么，这是否意味着 longLived 属性仍然被绑定，但它永远不会被更新？这是否使 longLived 属性处于绑定状态(使进一步绑定变得不可能)，但什么也不做？

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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，则 shortLivedProperty 将永远不会再次失效。因此，longLived 的侦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?

无论绑定状态如何，您都应该始终能够 bind 将属性添加到新的 observable，因为旧的 observable 属性将为您删除/解除绑定：

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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();
    }
}
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