Question

我正在尝试创建一个将UIViewController作为函数的函数。原因是有多个自定义视图控制器可以传递。这是我当前的函数，它可以使用switch语句和enum：

enum controllerTypes {
    case First, Second
}

extension UIViewController {

    func presentViewController(storyBoardName: String, storyBoardIdentifier: String, controllerType: controllerTypes, completion:(() -> Void)?) {

        switch controllerType {

        case .First:
            let firstVC = UIStoryboard(name: storyBoardName, bundle: nil).instantiateViewController(withIdentifier: storyBoardIdentifier) as? FirstViewController
            if let firVC = firstVC {
                self.present(firVC, animated: true, completion: nil)
            }
        case .Second:
            let secondVC = UIStoryboard(name: storyBoardName, bundle: nil).instantiateViewController(withIdentifier: storyBoardIdentifier) as? SecondViewController
            if let secVC = secondVC {
                self.present(secVC, animated: true, completion: nil)
            }
        }
        completion?()
    }
}

而不是传递'controllerTypes'枚举的参数我想传递任何类型的UIViewController，当我尝试这样做时，我得到以下错误：

        func presentViewController(storyBoardName: String, storyBoardIdentifier: String, controllerType: UIViewController, completion:(() -> Void)?) {
            let sampleVC = UIStoryboard(name: storyBoardName, bundle: nil).instantiateViewController(withIdentifier: storyBoardIdentifier) as? controllerType//error - use of undeclared type 'controllerType'
            if let samVC = sampleVC {
                self.present(samVC, animated: true, completion: nil)
            }
}

是否可以这样做？

Answer 1

您必须使您的函数通用，然后转换为通用参数类型：

extension UIViewController {
    func presentViewController<T: UIViewController>(storyBoardName: String, storyBoardIdentifier: String, controllerType: T.Type, completion:(() -> Void)?) {
        let sampleVC = UIStoryboard(name: storyBoardName, bundle: nil).instantiateViewController(withIdentifier: storyBoardIdentifier) as? T
        if let samVC = sampleVC {
            self.present(samVC, animated: true, completion: completion)
        }
    }
}

创建一个将UIViewController作为参数

1 个答案: