我可以通过快速扩展添加协议一致的协议吗？
//Plain old protocol here
protocol MyData {
var myDataID: Int { get }
}
我想使默认的MyData协议相当(只是比较ID)
extension MyData : Equatable { }
但是我得到这个可爱的错误：
“Extension of protocol ‘MyData’ cannot have an inheritance clause”
我正在寻找的行为是符合Equatable(协议)的BananaData,因为它实现了MyData协议,该协议可以提供Equatable
//This is the method to implement Equatable
func ==(lhs: MyData,rhs: MyData) -> Bool {
return lhs.myDataID == rhs.myDataID
}
struct BananaData: MyData {
var myDataID: Int = 1
}
func checkEquatable(bananaOne: BananaData,bananaTwo: BananaData) {
//This compiles,verifying that BananaData can be compared
if bananaOne == bananaTwo { }
//But BananaData is not convertible to Equatable,which is what I want
let equatableBanana = bananaOne as Equatable
//I don't get the additional operations added to Equatable (!=)
if bananaOne != bananaTwo { } //Error
}
由于错误消息说：协议的扩展不能有继承子句.相反,您可以使MyData协议在原始声明中从Equatable继承.
原文链接：https://www.f2er.com/swift/319810.htmlprotocol MyData: Equatable {
var myDataID: Int { get }
}
然后,可以扩展为符合MyData的类型添加一个==的实现：
func == <T: MyData>(lhs: T,rhs: T) -> Bool {
return lhs.myDataID == rhs.myDataID
}
但是,我不会推荐这个！如果您添加更多属性到符合类型,则不会检查其属性是否相等.举个例子：
struct SomeData: MyData {
var myDataID: Int
var myOtherData: String
}
let b1 = SomeData(myDataID: 1,myOtherData: "String1")
let b2 = SomeData(myDataID: 1,myOtherData: "String2")
b1 == b2 // true,although `myOtherData` properties aren't equal.
在上述情况下,您需要为SomeData重写==以获得正确的结果,从而使得接受MyData冗余的==.
