Question

与Swift 3.1 deprecates initialize(). How can I achieve the same thing?遇到同样的问题，@ Jordan Smith的解决方案非常令人印象深刻，然后我对这个工具很感兴趣，但在实践中遇到了一些麻烦，这里有一些关键代码，请看评论，为什么UIViewController中的日志功能未被调用，它符合协议;为什么UIViewController被抓住了，T == UIViewController.self是false：

protocol Conscious {
    static func awake()
}

/** extension */
extension UIViewController: Conscious {
    static func awake() {
        if self == UIViewController.self {
            print(self, #function)     // never came here, but seems should come
        }
    }
}

/** main */
private static let _operation: Void = {
    let typeCount = Int(objc_getClassList(nil, 0))
    let types = UnsafeMutablePointer<AnyClass?>.allocate(capacity: typeCount)
    let autoreleasingTypes = AutoreleasingUnsafeMutablePointer<AnyClass?>(types)
    objc_getClassList(autoreleasingTypes, Int32(typeCount))

    for index in 0 ..< typeCount {
        (types[index] as? Conscious.Type)?.awake()

        let T = types[index]!
        let vc = UIViewController()
        print(T, vc.isKind(of: T), T == UIViewController.self)
        /*
          Strange things:
          UIResponder true false
          UIViewController true false(why is false)
          UISearchController false false
         */
    }

    types.deallocate(capacity: typeCount)
}()

Answer 1

好吧，看起来使用objc方法看不到UIViewController的swift扩展。所以这是基于this topic的解决方案。

首先，您需要使用Conscious关键字标记@objc - 协议。

@objc protocol Conscious {
    static func awake()
}

然后你需要使用class_conformsToProtocol - 函数。

let type = types[index]!
if class_conformsToProtocol(type, Conscious.self) {
    let item = type as! Conscious.Type
    item.awake()
}

对swift中的类型判断感到困惑？

1 个答案: