ios 为什么我不能在 Swift 中调用 UIViewController 上的默认 super.init()？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/24095037/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Why can't I call the default super.init() on UIViewController in Swift?
提问by SirRupertIII
I am not using a UIViewControllerfrom a storyboard and I want to have a custom initfunction where I pass in an NSManagedObjectIDof some object. I just want to call super.init()like I have in Objective-C. Like this:
我没有使用UIViewController故事板中的 a ，我想要一个自定义init函数，我可以在其中传入NSManagedObjectID某个对象的 。我只想像super.init()在 Objective-C 中那样调用。像这样：
init(objectId: NSManagedObjectID) {
super.init()
}
But I get the following compiler error:
但我收到以下编译器错误：
Must call designated initializer of the superclass UIViewController
必须调用超类 UIViewController 的指定构造器
Can I simply not do this anymore?
我可以不再这样做了吗？
回答by occulus
The designated initialiser for UIViewControlleris initWithNibName:bundle:. You should be calling that instead.
的指定初始值设定项UIViewController是initWithNibName:bundle:。你应该打电话给那个。
If you don't have a nib, pass in nilfor the nibName (bundle is optional too). Then you could construct a custom view in loadViewor by adding subviews to self.viewin viewDidLoad, same as you used to.
如果您没有笔尖，请输入笔尖nil名称（捆绑也是可选的）。然后，您可以在 in 中loadView或通过向self.viewin添加子视图来构建自定义视图viewDidLoad，就像以前一样。
回答by Klaas
Another nice solution is to declare your new initializer as a convenienceinitializer as follows:
另一个不错的解决方案是将您的新初始化程序声明为convenience初始化程序，如下所示：
convenience init( objectId : NSManagedObjectID ) {
self.init()
// ... store or user your objectId
}
If you declare no designated initializers in your subclass at all, they are inherited automatically and you are able to use self.init()within your convenience initializer.
如果您在子类中根本没有声明任何指定的构造器，它们将被自动继承，并且您可以self.init()在您的便利构造器中使用它们。
In case of UIViewController the default init method will call init(nibName nibNameOrNil: String!, bundle nibBundleOrNil: NSBundle!)with nilfor both arguments (Command-Click on UIViewController will give you that info).
在UIViewController中的情况下，默认的init（）方法将调用init(nibName nibNameOrNil: String!, bundle nibBundleOrNil: NSBundle!)具有nil两个参数（命令点击UIViewController中会给你的信息）。
TL;TR: If you prefer to programmatically work with UIViewControllers here is a complete working example that adds a new initializer with a custom argument:
TL;TR：如果您更喜欢以编程方式使用UIViewControllers 这里是一个完整的工作示例，它添加了一个带有自定义参数的新初始化程序：
class MyCustomViewController: UIViewController {
var myString: String = ""
convenience init( myString: String ) {
self.init()
self.myString = myString
}
}
回答by NcNc
Update: add the link
更新：添加链接
https://developer.apple.com/documentation/uikit/uiviewcontroller/1621359-init
https://developer.apple.com/documentation/uikit/uiviewcontroller/1621359-init
According to the documentation for iOS, the designated initialiser for UIViewController is initWithNibName: bundle:.
根据 iOS 的文档， UIViewController 的指定初始化程序是initWithNibName: bundle:.
If you subclass UIViewController, you must call the super implementation of this method, even if you aren't using a NIB.
如果你继承 UIViewController，你必须调用这个方法的超级实现，即使你没有使用 NIB。
You can do it as follows:
你可以这样做：
init(objectId : NSManagedObjectID) {
super.init(nibName: (xib's name or nil'), bundle: nil)
// other code...
}
or
或者
Declare a new initializer as a convenience initializer:
声明一个新的初始化器作为便利初始化器：
convenience init( objectId : NSManagedObjectID ) {
self.init()
// other code...
}
}
回答by Vyacheslav
To improve the occulus'sanswer:
改进occulus 的答案：
init() {
super.init(nibName: nil, bundle: nil)
}
