ResponseSerializer'无法使用Swift 3调用非函数类型'NSHTTPURLResponse？'的值

Question

在更新到Xcode 8 beta 6之前,我一直在使用以下代码而没有问题.它与Alamofire存储库中的此示例类似.今天早上我将我的Alamofire库更新到最新的swift3分支,现在与beta 6兼容.它显示错误:此处Cannot call value of non-function type 'HTTPURLResponse?'存在类似的问题,但它不是基于当前版本的Swift和Alamofire.

根据我的理解,这个错误是因为它认为我试图返回Request属性response而不是函数,response(responseSerializer: <T>, completionHandler: <(Response<T.SerializedObject, T.ErrorObject>) -> Void>)并且它认为这是因为我responseSerializer或者completionHandler传入函数中的类型错误.

如何调整此代码以使其与函数声明和编译器兼容？

我添加@escaping到completionHandler来更正错误.

import Foundation
import Alamofire
import SwiftyJSON

extension Alamofire.Request {
public func responseObject<T: ResponseJSONObjectSerializable>(_ completionHandler: @escaping (Response<T, NSError>) -> Void) -> Self {
    let responseSerializer = ResponseSerializer<T, NSError> { request, res, data, error in

        guard let responseData = data else {
            let error = DFError.error(withDFCode: .dataSerializationFailed, failureReason: "Data could not be serialized because input data was nil.")
            return .failure(error)
        }

        let jsonData: Any?
        do {
            jsonData = try JSONSerialization.jsonObject(with: responseData, options: [])
        } catch  {
            let error = DFError.error(withDFCode: .jsonSerializationFailed, failureReason: "JSON could not be serialized into response object")
            return .failure(error)
        }

        let json = SwiftyJSON.JSON(jsonData!)
        if let newObject = T(json: json) {

            return .success(newObject)
        }

        let error = DFError.error(withDFCode: .jsonSerializationFailed, failureReason: "JSON could not be serialized into response object")
        return .failure(error)
    }

    return response(responseSerializer: responseSerializer, completionHandler: completionHandler)
    //Error: Cannot call value of non-function type 'HTTPURLResponse?'
}
}

Answer 1

你需要标记你completionHandler的@escaping.