延迟初始化并保留周期

本文介绍了延迟初始化并保留周期的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

使用惰性初始化程序时，是否有保留周期的机会?

While using lazy initialisers, is there a chance of having retain cycles?

在博客文章和许多其他地方[unowned self]被看到

In a blog post and many other places [unowned self] is seen

class Person { var name: String lazy var personalizedGreeting: String = { [unowned self] in return "Hello, \(self.name)!" }() init(name: String) { self.name = name } }

我尝试过

class Person { var name: String lazy var personalizedGreeting: String = { //[unowned self] in return "Hello, \(self.name)!" }() init(name: String) { print("person init") self.name = name } deinit { print("person deinit") } }

像这样使用它

//... let person = Person(name: "name") print(person.personalizedGreeting) //..

并发现已记录人员deinit".

And found that "person deinit" was logged.

因此，似乎没有保留周期. 据我所知，当一个块捕获自身时，以及当该块被自身强烈保留时，会有一个保留周期.这种情况似乎类似于保留周期，但实际上并非如此.

So it seems there are no retain cycles. As per my knowledge when a block captures self and when this block is strongly retained by self, there is a retain cycle. This case seems similar to a retain cycle but actually it is not.

推荐答案

我尝试过此[...]

I tried this [...]

lazy var personalizedGreeting: String = { return self.name }()

似乎没有保留周期

it seems there are no retain cycles

正确.

原因是，立即应用的闭包{}()被视为@noescape.它不会保留捕获的self.

The reason is that the immediately applied closure {}() is considered @noescape. It does not retain the captured self.

供参考:乔·格罗夫的推文.