我有以下课程:
class BaseCache<T: Equatable>: NSObject {
var allEntities = [T]()
// MARK: - Append
func appendEntities(newEntities: [T]) {
....
}
}
现在我想继承它,但是我得到了令人讨厌的错误,我的类型&#34;不符合协议&#39; Equatable&#39;&#34;:
看起来斯威夫特的仿制品真的很痛苦。
答案 0 :(得分:16)
您对TrackingCache
的类定义是错误的。它重复通用参数:
class TrackingCache<AftershipTracking>: BaseCache<AftershipTracking> { }
应该遗漏:
class TrackingCache: BaseCache<AftershipTracking> { }
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> { }
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
答案 1 :(得分:1)
这应该有效:&lt; swift 4&gt;
class TrackingCache<T: AftershipTracking>: BaseCache<T>
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
c
现在是CS
。