在 Swift 中使用具有约束关联类型的协议作为属性

Question

我正在尝试实现具有关联类型的数据源协议

protocol DataSourceCompatible {
    associatedtype CellModel
    func cellModelForItem(at indexPath: IndexPath) -> CellModel
}

协议AddressBookViewModelType 继承自基本协议并将关联的值约束到另一个协议

protocol AddressBookViewModelType: class, DataSourceCompatible where CellModel == AddressBookCellModelType {
}

AddressBookViewModel是AddressBookViewModelType协议的具体实现

class AddressBookViewModel: AddressBookViewModelType {
    func cellModelForItem(at indexPath: IndexPath) -> AddressBookCellModelType {
        let contact = sectionedContacts[indexPath.section][indexPath.row]
        return AddressBookCellModel(contact: contact)
    }
}

代码编译得很好，但是当我将视图模型声明为视图控制器上的属性时，编译器会失败并显示Protocol 'AddressBookViewModelType' can only be used as a generic constraint because it has Self or associated type requirements。

class AddressBookViewController: UIViewController {
    private var viewModel: AddressBookViewModelType!

    func configure(viewModel: AddressBookViewModelType) {
        self.viewModel = viewModel
    }
...
}

我记得看到类型擦除可能会解决问题，但我对类型擦除概念并不熟悉。有没有办法解决这个问题？

更新：

AddressBookCellModelType 和 AddressBookCellModel 在这里是如何关联的？

这是一个实现协议的结构体。

protocol AddressBookCellModelType {
    var name: String { get }
    var photo: UIImage? { get }
    var isInvited: Bool { get }
}

struct AddressBookCellModel: AddressBookCellModelType {
 ....
}

Answer 1

您是否尝试过仅将其用作泛型，如警告/错误所说：

class AddressBookViewController<T: AddressBookViewModelType> : UIViewController {
    private var viewModel: T!

    func configure(viewModel: T) {
        self.viewModel = viewModel
    }
    ...
}

您需要使用变量T 的属性初始化控制器，以便推断类型。

Answer 2

为了扩展我在 cmets 中的问题，看看这段代码，它看起来就像不添加 AddressBookCellModelType 或 AddressBookViewModelType 一样灵活，这也可以摆脱头痛，同时仍然是通用的DataSourceCompatible.

// This protocol is fine and very useful for making reusable view controllers. Love it.
protocol DataSourceCompatible {
    associatedtype CellModel
    func cellModelForItem(at indexPath: IndexPath) -> CellModel
}

// No need for a protocol here. The struct is its own interface.
// This ensures value semantics, which were being lost behind the protocol
// (since a protocol does not promise value semantics)    
struct AddressBookCellModel {
    var name: String
    var photo: UIImage?
    var isInvited: Bool
}

// AddressBookViewModel conforms to DataSourceCompatible
// Its conformance sets CellModel to AddressBookCellModel without needing an extra protocol
class AddressBookViewModel: DataSourceCompatible {
    let sectionedContacts: [[AddressBookCellModel]] = []
    func cellModelForItem(at indexPath: IndexPath) -> AddressBookCellModel {
        return sectionedContacts[indexPath.section][indexPath.row]
    }
}

class AddressBookViewController: UIViewController {
    private var viewModel: AddressBookViewModel!

    func configure(viewModel: AddressBookViewModel) {
        self.viewModel = viewModel
    }
}

这样做允许一个通用的 VC，而无需引入更多所需的部分：

class DataSourceViewController<DataSource: DataSourceCompatible>: UIView {
    private var viewModel: DataSource.CellModel!

    func configure(viewModel: DataSource.CellModel) {
        self.viewModel = viewModel
    }
}

let vc = DataSourceViewController<AddressBookViewModel>()

Answer 3

这只是 Swift 规范，您不能使用“具有关联类型的协议”作为类型声明。原因是编译器在编译时无法知道关联的类型实际上是什么，这违反了 Swift 的“类型安全”。

解决方案是像你说的那样使用类型橡皮擦，或者使类型通用。