swift - 我们可以在 Swift 中创建具有非可选属性的类型删除弱引用吗？

Question

一些背景

类型删除容器是 Swift 中的有用结构，因为它目前无法支持传递泛型类型参数。社区对此有一些很好的解释:

• http://www.russbishop.net/type-erasure
• https://realm.io/news/tryswift-gwendolyn-weston-type-erasure/
• https://www.bignerdranch.com/blog/breaking-down-type-erasures-in-swift/

这是一个例子:

protocol View: class {
    associatedtype ViewModel: Equatable

    var viewModel: ViewModel! { get set }

    func render(_ viewModel: ViewModel)
}

class _AnyViewBoxBase<T: Equatable>: View {

    var viewModel: T!

    func render(_ viewModel: T) {
        fatalError()
    }
}

final class _ViewBox<Base: View>: _AnyViewBoxBase<Base.ViewModel> {

    var base: Base!

    override var viewModel: Base.ViewModel! {
        get {
            return base.viewModel
        }
        set {
            base.viewModel = newValue
        }
    }

    init(_ base: Base) {
        self.base = base
    }

    override func render(_ viewModel: Base.ViewModel) {
        base.render(viewModel)
    }
}

final class AnyView<T: Equatable>: View {

    var _box: _AnyViewBoxBase<T>

    var viewModel: T! {
        get {
            return _box.viewModel
        }
        set {
            _box.viewModel = newValue
        }
    }

    func render(_ viewModel: T) {
        _box.render(viewModel)
    }

    init<Base: View>(_ base: Base) where Base.ViewModel == T {
        _box = _ViewBox(base)
    }
}

struct ExampleViewModel {
    let content: String
}

extension ExampleViewModel: Equatable {
    static func ==(lhs: ExampleViewModel, rhs: ExampleViewModel) -> Bool {
        return lhs.content == rhs.content
    }
}

final class Example: View {
    var viewModel: ExampleViewModel!

    init(viewModel: ExampleViewModel) {
        self.viewModel = viewModel
    }

    func render(_ viewModel: ExampleViewModel) {
    }
}

这些类型删除盒允许我们构建通用容器或创建必须符合具有特定类型的通用协议(protocol)但不限于具体实现的属性。例如使用 AnyView下面我可以轻松地交换一个 View 测试替身。

struct TypeUnderTest {
    var view: AnyView<ExampleViewModel>
}

var example = Example(viewModel: ExampleViewModel(content: "hello"))
var instanceUnderTest = TypeUnderTest(view: AnyView(example))

---

到目前为止一切顺利。我可以类似地定义 View有一个可选的或非可选的(而不是隐式解包可选)viewModel属性并相应地按框更新。

但是，如果我希望我的类型删除属性成为一个弱引用怎么办？

weak var view: AnyView<ExampleViewModel>不好。这将使我只剩下对盒子类型的弱引用，并且它会立即被释放。

var view: WeakAnyView<ExampleViewModel>让我们更接近。我们可以创建一个弱引用其内容的框。如果我们的View协议(protocol)只定义可选属性然后我们就可以开始了:

protocol View: class {
    associatedtype ViewModel: Equatable

    var viewModel: ViewModel? { get set }

    func render(_ viewModel: ViewModel)
}

class _AnyViewBoxBase<T: Equatable>: View {

    var viewModel: T?

    func render(_ viewModel: T) {
        fatalError()
    }
}

final class _ViewBox<Base: View>: _AnyViewBoxBase<Base.ViewModel> {

    weak var base: Base?

    override var viewModel: Base.ViewModel? {
        get {
            return base?.viewModel
        }
        set {
            base?.viewModel = newValue
        }
    }

    init(_ base: Base) {
        self.base = base
    }

    override func render(_ viewModel: Base.ViewModel) {
        base?.render(viewModel)
    }
}

final class AnyView<T: Equatable>: View {

    var _box: _AnyViewBoxBase<T>

    var viewModel: T? {
        get {
            return _box.viewModel
        }
        set {
            _box.viewModel = newValue
        }
    }

    func render(_ viewModel: T) {
        _box.render(viewModel)
    }

    init<Base: View>(_ base: Base) where Base.ViewModel == T {
        _box = _ViewBox(base)
    }
}

struct ExampleViewModel {
    let content: String
}

extension ExampleViewModel: Equatable {
    static func ==(lhs: ExampleViewModel, rhs: ExampleViewModel) -> Bool {
        return lhs.content == rhs.content
    }
}

final class Example: View {
    var viewModel: ExampleViewModel?

    init(viewModel: ExampleViewModel?) {
        self.viewModel = viewModel
    }

    func render(_ viewModel: ExampleViewModel) {
    }
}

struct TypeUnderTest {
    var view: AnyView<ExampleViewModel>
}

let viewModel = ExampleViewModel(content: "hello")
var example: Example? = Example(viewModel: viewModel)
let instanceUnderTest = TypeUnderTest(view: AnyView(example!))
instanceUnderTest.view.viewModel
example = nil
instanceUnderTest.view.viewModel

---

但是，如果我删除的协议(protocol) (View) 定义了非可选属性，那么我们就会遇到问题。 _ViewBox必须定义一个非可选的 viewModel符合 View但这迫使我们忽略了一个非常现实的可能性，即我们的弱引用装箱类型将被释放，并且我们没有安全的方式将其传达给调用者。

一个选择是添加另一个框，但这使用起来会变得很痛苦:

protocol View: class {
    associatedtype ViewModel: Equatable

    var viewModel: ViewModel { get set }

    func render(_ viewModel: ViewModel)
}

class _AnyViewBoxBase<T: Equatable>: View {

    var viewModel: T

    func render(_ viewModel: T) {
        fatalError()
    }

    init(viewModel: T) {
        self.viewModel = viewModel
    }

    var empty: Bool {
        get {
            return false
        }
    }
}

final class _ViewBox<Base: View>: _AnyViewBoxBase<Base.ViewModel> {

    weak var base: Base?

    override var viewModel: Base.ViewModel {
        get {
            return base!.viewModel
        }
        set {
            base?.viewModel = newValue
        }
    }

    init(_ base: Base) {
        super.init(viewModel: base.viewModel)
        self.base = base
    }

    override func render(_ viewModel: Base.ViewModel) {
        base?.render(viewModel)
    }

    override var empty: Bool {
        get {
            return base == nil
        }
    }
}

final class AnyView<T: Equatable>: View {

    var _box: _AnyViewBoxBase<T>

    var viewModel: T {
        get {
            return _box.viewModel
        }
        set {
            _box.viewModel = newValue
        }
    }

    func render(_ viewModel: T) {
        _box.render(viewModel)
    }

    init<Base: View>(_ base: Base) where Base.ViewModel == T {
        _box = _ViewBox(base)
    }

    var empty: Bool {
        return _box.empty
    }
}

struct AnyViewOptionalBox<T: Equatable> {

    private var _view: AnyView<T>?
    var view: AnyView<T>? {
        get {
            if let view = self._view, view.empty == false {
                return view
            } else {
                return nil
            }
        }
        set {
            self._view = newValue
        }
    }

    init(view: AnyView<T>) {
        self.view = view
    }
}

struct ExampleViewModel {
    let content: String
}

extension ExampleViewModel: Equatable {
    static func ==(lhs: ExampleViewModel, rhs: ExampleViewModel) -> Bool {
        return lhs.content == rhs.content
    }
}

final class Example: View {
    var viewModel: ExampleViewModel

    init(viewModel: ExampleViewModel) {
        self.viewModel = viewModel
    }

    func render(_ viewModel: ExampleViewModel) {
    }
}

struct TypeUnderTest {
    var viewBox: AnyViewOptionalBox<ExampleViewModel>
}

let viewModel = ExampleViewModel(content: "hello")
var example: Example? = Example(viewModel: viewModel)
let anyView: AnyView<ExampleViewModel> = AnyView(example!)
let anyViewOptional: AnyViewOptionalBox<ExampleViewModel> = AnyViewOptionalBox(view: anyView)
let instanceUnderTest = TypeUnderTest(viewBox: anyViewOptional)
instanceUnderTest.viewBox.view?.viewModel.content
example = nil
instanceUnderTest.viewBox.view?.viewModel.content

---

有没有更好的方法来维护对类型删除属性的弱引用？

Answer 1

基本上您想要的是将类型删除框的生命周期链接到它包含的对象的生命周期，这样一旦包含的对象被释放，盒子就会被释放。

一种方法是确保盒子仅弱引用包含的对象，并使用 objc_setAssociatedObject(...) 使盒子成为包含对象的关联对象。这样，您基本上颠倒了两个对象之间的所有权关系。

请参阅下面的 playground 示例:

import ObjectiveC

protocol View: class {
    associatedtype ViewModel: Equatable

    var viewModel: ViewModel { get set }

    func render()
}

private var AssociatedObjectHandle: UInt8 = 0

final class AnyView<T: Equatable>: View {

    let _viewModelGetter: () -> T
    let _viewModelSetter: (T) -> Void
    let _render: () -> Void

    init<Base: View>(_ base: Base) where Base.ViewModel == T {
        //Ensure this object doesn't reference base, so there is no retain cycle
        _viewModelGetter = { [weak base] in
            //You can force unwrap, because it is guaranteed that base is not deallocated because of the association
            return base!.viewModel
        }
        _viewModelSetter = { [weak base] in
            //You can force unwrap, because it is guaranteed that base is not deallocated because of the association
            base!.viewModel = $0
        }
        _render = { [weak base] in
            //You can force unwrap, because it is guaranteed that base is not deallocated because of the association
            base!.render()
        }

        //Associate this object with the base, so it gets deallocated when base gets deallocated, also base is guaranteed to exist during our lifetime
        objc_setAssociatedObject(base, &AssociatedObjectHandle, self, objc_AssociationPolicy.OBJC_ASSOCIATION_RETAIN_NONATOMIC)
    }

    deinit {
        print("dealloc: \(self)")
    }

    var viewModel: T {
        get {
            return _viewModelGetter()
        }
        set {
            _viewModelSetter(newValue)
        }
    }

    func render() {
        _render()
    }
}

class ConcreteView: View {
    typealias ViewModel = String

    var viewModel: String

    init(viewModel: String) {
        self.viewModel = viewModel
    }

    deinit {
        print("dealloc: \(self)")
    }

    func render() {
        print("viewModel: \(viewModel)")
    }
}


weak var anyView: AnyView<String>?
autoreleasepool {
    var concreteView = ConcreteView(viewModel: "Test")
    autoreleasepool {
        anyView = AnyView(concreteView)

        //Any view should render correctly because concrete view exists
        anyView!.render()
    }
    //Success: anyView is not nil yet, because concreteView still exists
    anyView!.render()
}
//Crash: anyView is now nil
anyView!.render()

输出:

viewModel: Test
viewModel: Test
dealloc: __lldb_expr_34.ConcreteView
dealloc: __lldb_expr_34.AnyView<Swift.String>
Fatal error: Unexpectedly found nil while unwrapping an Optional value