在Swift中，为什么子类方法不能覆盖超类的协议扩展所提供的方法 [英] In Swift, why subclass method cannot override the one, provided by protocol extension in superclass

问题描述

我知道这个问题的标题令人困惑，但以下示例说明了奇怪的行为:

I know the title of this question is confusing but the weird behaviour is explained in the example below:

protocol Protocol {
    func method() -> String
}

extension Protocol {
    func method() -> String {
        return "From Base"
    }
}

class SuperClass: Protocol {
}

class SubClass: SuperClass {
    func method() -> String {
        return "From Class2"
    }
}

let c1: Protocol = SuperClass()
c1.method() // "From Base"
let c2: Protocol = SubClass()
c2.method() // "From Base"

c1.method()和c2.method()为何返回相同的值?子类中的method()为什么不起作用?

How come c1.method() and c2.method() return the same? How come the method() in SubClass doesn't work?

有趣的是，在没有声明c2类型的情况下，这将起作用:

Interestingly, without declaring the type of c2, this is going to work:

let c2  = SubClass()
c2.method() // "From Class2"

推荐答案

问题是c1和c2的类型为Protocol，因为您已经以这种方式明确定义了它们的类型(请记住:协议是完全成熟的类型).这意味着，在调用method()时，Swift会调用Protocol.method.

The problem is that c1 and c2 are of type Protocol, as you've defined their type explicitly this way (remember: protocols are fully fledged types). This means, when calling method(), Swift calls Protocol.method.

如果您定义如下内容:

let c3 = SuperClass()

... c3的类型为SuperClass.由于SuperClass没有更具体的method()声明，因此在调用c3.method()时仍使用Protocol.method().

...c3 is of type SuperClass. As SuperClass has no more specific method() declaration, Protocol.method() is still used, when calling c3.method().

如果您定义如下内容:

let c4 = SubClass()

... c4的类型为SubClass.由于SubClass确实具有更具体的method()声明，因此在调用c4.method()时将使用SubClass.method().

...c4 is of type SubClass. As SubClass does have a more specific method() declaration, SubClass.method() is used, when calling c4.method().

您还可以通过将SubClass.method()向下广播到`SubClass:

You could also get c2 to call SubClass.method(), by down-casting it to `SubClass:

(c2 as! SubClass).method() // returns "From Class2"

---

这是 SwiftStub 上的演示.

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Here's a demonstration on SwiftStub.

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