Swift

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!
  }
}

我有一堆这样的，所以我想把它弄得更干净：

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 {
}

但我不知道如何在静态/class 函数中动态引用对象的类型，以及如何在从子class。一旦用 type(of: object)

实例化了一个对象，这很容易做到

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 100% 确定它是否有效，因为我没有直接测试你的代码（不想创建一个故事板只是为了测试），但我写了类似的代码来测试它，它有效：

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

Swift - 在静态/class 函数中引用当前 class 类型

Swift - Refer to current class type in a static / class function

static

class

subclass

instantiation