Swift泛型类实现泛型函数时类型转换错误求助

错误信息 #

Cannot convert return expression of type 'Model' to return type 'Model'

问题代码 #

泛型类代码 #

public class URLCacheAdapter<Model>:LoaderProtocol{

    public let adaptee:URLCache
    public let mapper: (Data,URLResponse) throws -> Model
    
    public typealias Mapper = (Data,URLResponse) throws -> Model
    
    public init(adaptee: URLCache,
                mapper: @escaping (Data,URLResponse) throws -> Model) {
        self.adaptee = adaptee
        self.mapper = mapper
    }
    
    public enum CacheErrors:Swift.Error{
        case dataNotFound
    }
    
    public func load<Model>() async throws -> Model {
        guard let cachedResponse = adaptee.cachedResponse(for: requet) else{
            throw CacheErrors.dataNotFound
        }
        return mapper(cachedResponse.data,cachedResponse.response) // Error here
    }
}

协议代码 #

public protocol LoaderProtocol{
    func load<T>() async throws -> T
}

泛型映射器代码 #

public class DataMapper{

    public enum Errors: Error{
        case invalidData
    }

    public static func map(_ data:Data,from response:HTTPURLResponse) throws -> People{
        guard response.statusCode == 200, let root = try? JSONDecoder().decode(PeopleModel.self, from: data) else{
            throw Errors.invalidData
        }
        return ConvertModelToDto.convertToPeopleData(model: root)
    }
}

问题原因 #

1. 泛型重名冲突：类的泛型参数Model和load函数的泛型参数Model是完全独立的两个类型。类的Model由初始化时传入的mapper决定，而函数的Model是调用时指定的任意类型，编译器无法将两者关联，导致返回值类型不匹配。
2. 协议与类的泛型不兼容：原LoaderProtocol要求load函数能返回任意类型<T>，但URLCacheAdapter类只能返回固定的Model类型，违背了协议要求，进一步加剧类型冲突。
3. 拼写错误：代码中requet是request的拼写错误，会导致未定义变量问题。

解决方案 #

1. 修改协议为泛型协议 #

让协议关联具体类型，匹配类的泛型：

public protocol LoaderProtocol {
    associatedtype Model
    func load() async throws -> Model
}

2. 修正泛型类的load函数 #

移除函数的泛型参数，直接使用类级别的泛型Model，同时修正拼写错误：

public class URLCacheAdapter<Model>: LoaderProtocol {

    public let adaptee: URLCache
    public let mapper: (Data, URLResponse) throws -> Model
    public let request: URLRequest
    
    public typealias Mapper = (Data, URLResponse) throws -> Model
    
    public init(adaptee: URLCache,
                request: URLRequest,
                mapper: @escaping (Data, URLResponse) throws -> Model) {
        self.adaptee = adaptee
        self.request = request
        self.mapper = mapper
    }
    
    public enum CacheErrors: Swift.Error {
        case dataNotFound
    }
    
    public func load() async throws -> Model {
        guard let cachedResponse = adaptee.cachedResponse(for: request) else {
            throw CacheErrors.dataNotFound
        }
        return try mapper(cachedResponse.data, cachedResponse.response)
    }
}

3. 映射器使用示例 #

DataMapper的map函数返回固定People类型，初始化URLCacheAdapter时直接传入即可：

let cache = URLCache.shared
let request = URLRequest(url: URL(string: "your-target-url")!)
let adapter = URLCacheAdapter<People>(adaptee: cache, request: request) { data, response in
    guard let httpResponse = response as? HTTPURLResponse else {
        throw DataMapper.Errors.invalidData
    }
    return try DataMapper.map(data, from: httpResponse)
}

内容的提问来源于stack exchange，提问作者Tushar Sharma