如何返回不透明类型的泛型类

Question

如何为泛型类返回不透明类型？

func makeFAQController() -> UIHostingController<some View> {
let viewModel = FAQViewModel()
let view = FAQView().environmentObject(viewModel)
let controller = UIHostingController(rootView: view)
return controller
}

错误：无法转换类型为'UIHostingController<某些视图>'以键入'UIHostingController<FAQView>'强制

func makeFAQController() -> UIHostingController<FAQView> {
let viewModel = FAQViewModel()
let view = FAQView().environmentObject(viewModel)
let controller = UIHostingController(rootView: view)
return controller
}

无法转换'UIHostingController<类型的值；某些视图>'到类型'UIHostingController<FAQView>'强制

如果可能的话，我希望避免类型擦除。

Answer 1

您应该返回基类(没有比base类更通用的了，可能只有元NSObject(，所以

func makeFAQController() -> UIViewController {    // << here !!
let viewModel = FAQViewModel()
let view = FAQView().environmentObject(viewModel)
let controller = UIHostingController(rootView: view)
return controller
}

Answer 2

您还可以在此处使用类型擦除的AnyView：

func makeFAQController() -> UIHostingController<AnyView> {
let viewModel = FAQViewModel()
let view = FAQView().environmentObject(viewModel)
let controller = UIHostingController(rootView: AnyView(view))
return controller
}

避免AnyView的另一种方法是使makeFaqController方法的封闭类通用，并在其中存储一个闭包/viewBuilder，用作创建时的视图宿主：

class FAQMaker<U: View> {
fileprivate let _prepare: (FAQView) -> U
init(_ prepare: @escaping (FAQView) -> U) {
self._prepare = prepare
}

func makeFAQController() -> UIHostingController<U> {
let prepared = _prapare(FAQView())
return UIHostingController(rootView: prepared)
}
}

前一种方法适用于将SwiftUI视图封装到UICollectionReusableView中。