如何返回不透明类型的泛型类

Question

如何返回泛型类的不透明类型？

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

错误:无法在强制中将“UIHostingController”类型的值转换为“UIHostingController”类型

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

在强制中，

不能将“UIHostingController”类型的值转换为“UIHostingController”类型

如果可能的话，我想避免类型擦除。

Answer 1

您应该返回基类(没有比基类更通用的类，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

您最好在这里使用类型擦除的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方法的封闭类成为通用类，并在其中存储一个闭包/viewBuilder，以便在创建时用作视图宿主：

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)
    }

}

前一种方法可以很好地将SwiftUI视图封装到UICollectionReusableView中。