Question

有没有办法实例化 - 设置 - 推送使用这样的辅助方法，但同时避免向下转换？

func pushController(id: String, setup: (_ vc: UIViewController) -> ()) {
  if let vc = storyboard?.instantiateViewController(withIdentifier: id) {
    setup(vc)
    navigationController?.pushViewController(vc, animated: true)
  }
}

// usage
pushController(id: "Cars") { vc in
  (vc as! CarsVC).brand = "BMW"
}

// ...want to avoid downcasting
vc.brand = "BMW"

Answer 1

我认为你不能避免向下倾斜，但你可以减少痛苦：

func pushController<VC: UIViewController>(id: String, setup: (_ vc: VC) -> ()) {
  if let vc = storyboard?.instantiateViewController(withIdentifier: id) as? VC {
    setup(vc)
    navigationController?.pushViewController(vc, animated: true)
  }
}

// usage
pushController(id: "Cars") { (vc: CarsVC) in
  vc.brand = "BMW"
}

未经测试，因此可能存在小问题。

编辑：我应该注意，当ID与错误的类型一起使用时，这会无声地失败。您可能需要在else之后添加if来处理此问题。

Answer 2

我能想到的最优雅的解决方案是使用泛型，比如这个（游乐场） - 例如：

import UIKit

extension UIViewController {
    func pushController<T:UIViewController> (id: String, setup: (_ vc: T) -> ()) {
        if let vc = self.storyboard?.instantiateViewController(withIdentifier: id) as? T {
            setup(vc)
            self.navigationController?.pushViewController(vc, animated: true)
        }
    }

}

class ViewControllerA:UIViewController {}

class ViewControllerB:UIViewController {
    var bValue:Int = 0
}

let vcA = ViewControllerA();

vcA.pushController(id: "B") {
    (vc:ViewControllerB) in
    vc.title = "view controller b"
    vc.bValue = 42;
}

我更倾向于使用显式泛型类型调用pushController，但不幸的是Swift 3不支持这个：

vcA.pushController<ViewControllerB>(id: "B") { // Error: cannot explicitly specialize a generic function
    vc in
    vc.title = "view controller b"
    vc.bValue = 42;
}

用于推送UIViewController的Helper方法

2 个答案: