Swift 4

Question

请考虑以下设置：

protocol MyProcotol {
}

class MyModel: MyProcotol {
}

enum Result<T> {
    case success(value: T)
    case failure
}

class Test {
    func test<T: MyProcotol>(completion: (Result<T>) -> Void) {
        let model = MyModel()
        let result = Result.success(value: model)
        completion(result)
    }
}

为什么我不能拨打completion(result)？我收到此错误：

无法转换类型＆＃39;结果＆＃39;预期参数类型＆＃39;结果＆lt; _＆gt;＆＃39;

任何解决方法？

Answer 1

您在通用函数中使用非泛型具体类型MyModel，但不起作用。

你可以做这样的事情

class Test {
    func test<T: MyProcotol>(item: T, completion: (Result<T>) -> Void) {
        let result : Result<T> = .success(value: item)
        completion(result)
    }
}

Answer 2

您可以使用强制转换来转换潜在的通用值。

Swift 4

protocol MyProcotol {}

struct MyModel: MyProcotol {
    let x: Int
}

struct TheirModel: MyProcotol {
    let y: Int
}

enum Result<T> {
    case success(value: T)
    case failure

    var value: T? {
        switch self {
            case .success(let value): return value
            case .failure: return nil
        }
    }
}

struct Test {
    enum ModelType {
        case my, their
    }

    static func get<T: MyProcotol>(type: ModelType, completion: (Result<T>) -> Void) {
        let model: Any
        switch type {
            case .my: model = MyModel(x: 42)
            case .their: model = TheirModel(y: 19)
        }

        if let value = model as? T { completion(.success(value: value)) }
        else { completion(.failure) }
    }
}

Test.get(type: .my) { (result: Result<MyModel>) in
    guard let value = result.value else { return }
    print("here is MyModel \(value) with value: \(value.x)")
}

Test.get(type: .their) { (result: Result<TheirModel>) in
    guard let value = result.value else { return }
    print("here is TheirModel \(value) with value: \(value.y)")
}

Test.get(type: .their) { (value: Result<MyModel>) in
    print("here is failure? \(value)")
}

Swift泛型错误：无法转换类型＆＃39;类型<t>＆＃39;预期参数类型＆＃39;类型＆lt; _＆gt;＆＃39;

2 个答案:

Swift 4