在清除正确的模式中清理强引用？

问题描述

有几种资源（，，加上我看到它在任何地方使用）建议从的 deinit 中的 NotificationCenter 中删除​​观察者UIViewController ，例如：

There are several resources (blog, SO question, plus I've seen it used everywhere) that recommend removing an observer from the NotificationCenter in the deinit of the UIViewController, e.g.:

deinit {
    NotificationCenter.default.removeObserver(self, name: NSNotification.Name.UIKeyboardWillShow, object: nil)
    NotificationCenter.default.removeObserver(self, name: NSNotification.Name.UIKeyboardWillHide, object: nil)
}

现在根据我不必关心从 NotificationCenter 因为它使用了弱引用，我看到了与其他引用一样的模式。

Now while according to another blog entry I don't have to care about removing an observer from NotificationCenter since it uses weak references, I've seen the same pattern used with other references.

这个问题让我烦恼。根据官方：

The question that bugs me. According to official documentation:

这是否意味着如果仍然有强烈的类引用， deinit 不会被调用，因此呈现 deinit 引用清理无用？如果 NotificationCenter 中仍然存在对 viewController 的强引用，则 viewController 的 deinit 永远不会被调用，对吗？因此，删除 deinit 中强大的refenreces永远不会真正起作用。

Doesn't this mean that if there still is a strong reference to the class, deinit will not get called, thus rendering the deinit references cleanup useless? If there is still a strong reference to the viewController from the NotificationCenter, then viewController's deinit will never get called, right? So removing the strong refenreces in deinit can never really work.

我在这里遗漏了什么吗？

Am I missing something here?

推荐答案

此声明

的deinit中从NotificationCenter中删除观察者过去是真的。

was true in the past.

您的陈述

是正确的。

观察者持有弱引用到目标对象。

An observer holds a weak reference to the target object.

这解释了为什么即使有多个活跃的观察者也会调用一个对象的 deinit 。

This explain why the deinit of an object will be called even if there are multiple active observers.

这是必需的 iOS 9之前，以防止观察者调用解除分配对象的方法。

This was needed prior to iOS 9 to prevent an observer from invoking a method of a deallocated object.

然而，不再需要从macOS 10.11取消注册观察者iOS 9.0

However unregistering an observer is no longer needed from macOS 10.11 and iOS 9.0

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