Swift变量初始化 [英] Swift Variables Initialization

问题描述

我有一个关于快速初始化变量的问题.

I have a question about variables initialization in swift.

我有两种方法来初始化变量(作为Objective-C中类的属性").

I have two ways to initialize a variable (as "property" of a class in Objective-C).

其中哪个是最正确的?

class Class {

  var label: UILabel!

  init() { ... label = UILabel() ... }

}

或

class Class {

  var label = UILabel()

  init() { … }

}

推荐答案

实际上，您有5种初始化属性的方法.

Actually you have 5 ways to initialize properties.

没有正确的方法，方法取决于需求.
基本上，尽可能将诸如 UILabel 之类的对象始终声明为常量( let ).

There is no correct way, the way depends on the needs.
Basically declare objects like UILabel always – if possible – as constant (let).

5种方法是:

• 声明行中的初始化

• Initialization in the declaration line

let label = UILabel(frame:...

在 init 方法中的初始化，您不必将属性声明为隐式未包装的可选内容.

Initialization in the init method, you don't have to declare the property as implicit unwrapped optional.

let label: UILabel
init() { ... label = UILabel(frame:...) ... }

前两种方法实际上是相同的.

The first two ways are practically identical.

• 以类似 viewDidLoad 的方法进行初始化，在这种情况下，您必须将该属性声明为(隐式展开)可选，并且还声明为 var

• Initialization in a method like viewDidLoad, in this case you have to declare the property as (implicit unwrapped) optional and also as var

var label: UILabel!

on viewDidLoad()
 ...
 label = UILabel(frame:...)
}

使用闭包进行初始化以分配默认(计算)值.初始化类后，将立即调用该闭包，并且无法在闭包中使用该类的其他属性.

Initialization using a closure to assign a default (computed) value. The closure is called once when the class is initialized and it is not possible to use other properties of the class in the closure.

let label: UILabel = {
   let lbl = UILabel(frame:...)
   lbl.text = "Foo"
   return lbl
}()

使用闭包的惰性初始化.第一次访问该属性时，闭包被称为(一次)，您可以使用该类的其他属性.
该属性必须声明为 var

let labelText = "Bar"

lazy var label: UILabel = {
   let lbl = UILabel(frame:...)
   lbl.text = "Foo" + self.labelText
   return lbl
}()

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