我有一個存在于我的應用程式中的物件（比如一個用戶物件），我想發布它（因為某些 UI 依賴于它）。由于 Liskov 替換原則，我不想將它作為 EnvironmentObject 傳遞，所以我決定制作一個包裝器。像這樣的東西：
protocol HasUser: ObservableObject {
var user: String { get }
}
class HasUserWrapper: HasUser {
var user: String { userGetter() }
private let userGetter: () -> String
let objectWillChange: AnyPublisher<Void, Never>
init<UO: HasUser>(wrapping userObject: UO) {
self.objectWillChange = userObject
.objectWillChange
.map { _ in () }
.eraseToAnyPublisher()
self.userGetter = { userObject.user }
}
}
extension HasUser {
func eraseToHasUserWrapper() -> HasUserWrapper {
HasUserWrapper(wrapping: self)
}
}
class ConcreteHasUser: HasUser {
@Published var user: String = "john"
}
struct UserView {
@ObservedObject var hasUser: HasUserWrapper = ConcreteHasUser().eraseToHasUserWrapper()
var body: some View {
Text(hasUser.user)
}
}
現在，我想讓這個包裝器通用。所以我開始了這樣的事情：
protocol HasThing: ObservableObject {
associatedtype Thing
var thing: Thing { get }
}
class HasThingWrapper<Thing>: HasThing {
var thing: Thing { thingGetter() }
private let thingGetter: () -> Foo
let objectWillChange: AnyPublisher<Void, Never>
init<HasThingType: HasThing>(wrapping hasThing: HasThingType) {
self.objectWillChange = hasThing
.objectWillChange
.map { _ in () }
.eraseToAnyPublisher()
self.thingGetter = { hasThing. }
}
}
extension HasThing {
func eraseToThingWrapper() -> HasThingWrapper<Thing> {
HasThingWrapper(wrapping: self)
}
}
protocol HasString: HasThing {
var thing: String { get }
}
class StringHolder: HasString {
@Published var thing: String = "hello"
}
我如何告訴編譯器包裝器內的泛型型別 Thing 應該與 HasThing 協議中的關聯型別 Thing 匹配？
uj5u.com熱心網友回復：
如果我理解正確，您只需要在以下內容中添加一個通用約束init：
init<HasThingType: HasThing>(wrapping hasThing: HasThingType)
where HasThingType.Thing == Thing
{
...
// then you can do:
self.thingGetter = { hasThing.thing }
}
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