Question

我有静态函数来实例化看起来像

的视图控制器

class AController: UIViewController {
  static func instantiate() -> AController {
    let storyboard = UIStoryboard(name: "AController", bundle: nil)
    let controller = s.instantiateInitialViewController() as? AController
    return controller!
  }
}

class BController: UIViewController {
  static func instantiate() -> BController {
    let storyboard = UIStoryboard(name: "BController", bundle: nil)
    let controller = storyboard.instantiateInitialViewController() as? BController
    return controller!
  }
}

我有很多他们所以我想把它变得更清洁：

class MYViewController: UIViewController {
  class func instantiate() -> self.type {
    let storyboard = UIStoryboard(name: "\(self.type)", bundle: nil)
    let controller = storyboard.instantiateInitialViewController() as? self.type
    return controller!
  }
}

class AController: MYViewController {
}

class BController: MYViewController {
}

但我不知道动态引用static / class函数中对象的类型，以及如何从子类调用时引用它。一旦使用type(of: object)

实例化对象，就很容易做到

Answer 1

试试这个：

class MYViewController: UIViewController {
    class func instantiate() -> Self {
        let storyboard = UIStoryboard(name: "\(self)", bundle: nil)
        let controller = storyboard.instantiateInitialViewController() as! Self
        return controller
    }
}

我不是百分之百确定它是否有效，因为我没有直接测试你的代码（不想仅仅为了测试而创建一个故事板），但是我编写了类似的代码来测试它，它起作用：

class MYViewController: UIViewController {
    class func instantiate() -> Self {
        print("type: \(self)")
        return self.init()
    }
}

Swift - 在静态/类函数中引用当前类类型

1 个答案: