ios Swift 覆盖实例变量
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Swift override instance variables
提问by D?nu? Mihai Florian
I know this answer has already been posted in some other form here, but I'd like to understand more about overriding instance variables in swift.
我知道这个答案已经以其他形式发布在这里，但我想了解更多关于在 swift 中覆盖实例变量的信息。
Let's say I have this code
假设我有这个代码
class BaseView:UIView{
let someVariable:Int = 1
// do some work with someVariable
}
class ExtendedView:BaseView{
let someVariable:Int = 2
}
Ok. From what I read, the constant requires an override prefix. Other answers said that I should declare the setter and getter? Why? I really don't care about those two. I just need the value replaced. I can't really use the initoverride because I'm inheriting from UIView and this might be quite dangerous (i think).
好的。从我读到的内容来看，常量需要一个覆盖前缀。其他答案说我应该声明setter和getter？为什么？我真的不在乎那两个。我只需要替换值。我不能真正使用init覆盖，因为我是从 UIView 继承的，这可能非常危险（我认为）。
Any suggestions are welcomed.
欢迎任何建议。
回答by Grimxn
As you say, you cannot simply redefine a constant in a subclass (it is a constant, after all). The error you get is "Cannot override with a stored property". It does appear to be possible to override a var, however, when I change the let someVariableto var someVariableI get "ambiguous use of 'someVariable'" when I access it in the subclass (note - same thing happens whether I use overrideor not).
正如您所说，您不能简单地在子类中重新定义一个常量（毕竟它是一个常量）。您得到的错误是“无法用存储的属性覆盖”。似乎确实可以覆盖 a var，但是，当我将 更改为 时，当我在子类中访问它时let someVariable，var someVariable我得到“'someVariable' 的歧义使用”（注意 - 无论我是否使用，都会发生同样的事情override）。
The simplest solution is to use a getter. This is really a function, so you can happily override it, the backing variable will be managed for you, and if you don't supply a setter ... it will be constant for each class:
最简单的解决方案是使用吸气剂。这真的是一个函数，所以你可以愉快地覆盖它，后备变量将为你管理，如果你不提供一个 setter ......它对于每个类都是常量：
class BaseView: UIView {
var someVariable: Int { get { return 1 } }
// do some work with someVariable
}
class ExtendedView: BaseView {
override var someVariable: Int { get { return 2 } }
}
let a = BaseView()
a.someVariable // 1
let b = ExtendedView()
b.someVariable // 2
As commentator @user3633673 points out, if you onlyhave a getter (and not a setter), you can drop the get, but I left it in for clarity of the principle. Here's the same without it...
正如评论员@user3633673 指出的那样，如果您只有一个 getter（而不是一个 setter），您可以删除get，但为了清晰起见，我保留了它。没有它也一样......
class BaseView: UIView {
var someVariable: Int { return 1 }
// do some work with someVariable
}
class ExtendedView: BaseView {
override var someVariable: Int { return 2 }
}
let a = BaseView()
a.someVariable // 1
let b = ExtendedView()
b.someVariable // 2
... and, of course, in Swift 5, you can drop the return:
...当然，在 Swift 5 中，您可以删除return：
class BaseView: UIView {
var someVariable: Int { 1 }
}
class ExtendedView: BaseView {
override var someVariable: Int { 2 }
}
