And*_*eev 13 generics inheritance ios swift
我有以下课程:
class BaseCache<T: Equatable>: NSObject {
var allEntities = [T]()
// MARK: - Append
func appendEntities(newEntities: [T]) {
....
}
}
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现在我想继承它,但是我得到了令人讨厌的错误,我的类型"不符合协议'Equatable'":

看起来斯威夫特的仿制药真的很痛苦.
las*_*sej 17
你的类定义TrackingCache是错误的.它重复通用参数:
class TrackingCache<AftershipTracking>: BaseCache<AftershipTracking> { }
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应该省略:
class TrackingCache: BaseCache<AftershipTracking> { }
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这会触发潜在的快速错误Classes derived from generic classes must also be generic.您可以通过指定AftershipTracking所需或继承的类型参数来解决此问题:
class TrackingCache<T: AftershipTracking>: BaseCache<AftershipTracking> { }
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完整示例:
class BaseCache<T: Equatable>: NSObject {
var items: [T] = []
func appendItems( items: [T]) {
self.items += items
didAppendItems()
}
func didAppendItems() {} // for overriding
}
class AftershipTracking: NSObject {
var identifier: Int
init( identifier: Int) {
self.identifier = identifier
super.init()
}
}
extension AftershipTracking: Equatable { }
func ==( lhs: AftershipTracking, rhs: AftershipTracking) -> Bool {
return lhs.identifier == rhs.identifier
}
class TrackingCache<T: AftershipTracking>: BaseCache<AftershipTracking> {
override func didAppendItems() {
// do something
}
}
let a = TrackingCache<AftershipTracking>()
let b = TrackingCache<AftershipTracking>()
a.appendItems( [AftershipTracking( identifier: 1)])
b.appendItems( [AftershipTracking( identifier: 1)])
let result = a.items == b.items // true
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这应该有效:<swift 4>
class TrackingCache<T: AftershipTracking>: BaseCache<T>
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另一个例子:
protocol P {
}
class C: P {
}
class CS: C {
}
class L<T:P> {
let c: T
init(_ c: T) {
self.c = c
}
}
class LS<T:CS>:L<T> {
}
let i = LS(CS())
i.c
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c是CS现在。