IBOutlets和IBactions需要！到底 [英] IBOutlets and IBactions require ! in the end
问题描述
我今天试图从Obj-C开始到Swift，我正在阅读文档。我试图在Swift中创建一个简单的IBOutlet，它经常给我这些错误。
I tried to start and go from Obj-C to Swift today and I was reading the documentation. I tried to create an easy IBOutlet in Swift and it constantly gave me those errors.
View Controller没有初始化
必需init（编码器aDecoder：NSCoder）{
fatalError（init（编码器:)尚未实现）}
required init(coder aDecoder: NSCoder) { fatalError("init(coder:) has not been implemented") }
IBOutletproperty具有非可选类型'UILabel'
并且不断弹出以下代码：
and that constantly pops up with this code:
@IBOutlet var outputLabel：UILabel
但是当我添加一个！标记，它运行没有错误，如此
but when I add an ! mark, it's running without errors like so
@IBOutlet var outputLabel：UILabel！
同样的事情发生在IBActions ...
Same thing happens for IBActions...
推荐答案
首先要了解，什么是实际上！和？
First of all get to know, what is actually ! and ?
- 使用
？：如果将来该值变为零，那么您可以对此进行测试。 - 使用
！：如果将来真的不应该为零，但最初需要为零。
- Use
?: if the value can become nil in the future, so that you test for this. - Use
!: if it really shouldn't become nil in the future, but it needs to be nil initially.
@IBOutlet：
当您在Swift中声明一个插座时，编译器会自动将类型转换为弱隐式解包可选并为其指定初始值 nil 。
When you declare an outlet in Swift, the compiler automatically converts the type to a weak implicitly unwrapped optional and assigns it an initial value of nil.
实际上，编译器替换 @IBOutlet var name：键入 @IBOutlet weak var name：输入！ = nil 。
In effect, the compiler replaces @IBOutlet var name: Type with @IBOutlet weak var name: Type! = nil.
Xcode会更改它并强制限制声明 @IBOutlet 非选项类型变量，因此遵循 @IBOutlet 的两种声明都有效至日期。
Xcode would change it and Force restrict on declare @IBOutlet non option type variable , so following both kind of declaration for @IBOutlet is Valid till date.
@IBOutlet var outputLabel : UILabel!
@IBOutlet var priceLabel : UILabel?
但是，如果您控制拖动beta 4标签的插座，则会发生这种情况： / strong>
However, if you control drag an outlet for a label in beta 4 this happens:
@IBOutlet var priceLabel : UILabel! = nil
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