在 Swift 中返回实例类型

Question

我正在尝试进行此扩展：

extension UIViewController
{
    class func initialize(storyboardName: String, storyboardId: String) -> Self
    {
        let storyboad = UIStoryboard(name: storyboardName, bundle: nil)
        let controller = storyboad.instantiateViewControllerWithIdentifier(storyboardId) as! Self

        return controller
    }
}

但是我得到编译错误：

错误：无法将“UIViewController”类型的返回表达式转换为 返回类型 'Self'

可能吗？我也想把它做成

init(storyboardName: String, storyboardId: String)

Answer 1

与 在 Swift 中的类扩展函数中使用“self”类似，您可以定义一个通用辅助方法，该方法从调用上下文推断 self 的类型：

extension UIViewController
{
    class func instantiateFromStoryboard(storyboardName: String, storyboardId: String) -> Self
    {
        return instantiateFromStoryboardHelper(storyboardName, storyboardId: storyboardId)
    }

    private class func instantiateFromStoryboardHelper<T>(storyboardName: String, storyboardId: String) -> T
    {
        let storyboard = UIStoryboard(name: storyboardName, bundle: nil)
        let controller = storyboard.instantiateViewControllerWithIdentifier(storyboardId) as! T
        return controller
    }
}

然后

let vc = MyViewController.instantiateFromStoryboard("name", storyboardId: "id") 编译，类型推断为

MyViewController

。

---

更新

Swift 3：

extension UIViewController
{
    class func instantiateFromStoryboard(storyboardName: String, storyboardId: String) -> Self
    {
        return instantiateFromStoryboardHelper(storyboardName: storyboardName, storyboardId: storyboardId)
    }

    private class func instantiateFromStoryboardHelper<T>(storyboardName: String, storyboardId: String) -> T
    {
        let storyboard = UIStoryboard(name: storyboardName, bundle: nil)
        let controller = storyboard.instantiateViewController(withIdentifier: storyboardId) as! T
        return controller
    }
}

另一种可能的解决方案，使用

unsafeDowncast

：

extension UIViewController
{
    class func instantiateFromStoryboard(storyboardName: String, storyboardId: String) -> Self
    {
        let storyboard = UIStoryboard(name: storyboardName, bundle: nil)
        let controller = storyboard.instantiateViewController(withIdentifier: storyboardId)
        return unsafeDowncast(controller, to: self)
    }
}

Answer 2

Self

 是在编译时确定的，而不是运行时确定的。在您的代码中， 

Self

 完全等同于 

UIViewController

，而不是“恰好调用此函数的子类”。这将返回 

UIViewController

 并且调用者必须将 

as

 放入正确的子类中。我认为这就是您想要避免的（尽管这是“正常的可可”方式，所以仅返回 

UIViewController

 可能是最好的解决方案）。

注意：在任何情况下都不应该将函数命名为 initialize

。这是 

NSObject

 的现有类函数，最多会导致混乱，最坏的情况会导致错误。

但是如果你想避免调用者的

as

，子类化通常不是在 Swift 中添加功能的工具。相反，您通常需要泛型和协议。在这种情况下，您只需要泛型即可。

func instantiateViewController<VC: UIViewController>(storyboardName: String, storyboardId: String) -> VC {
    let storyboad = UIStoryboard(name name: storyboardName, bundle: nil)
    let controller = storyboad.instantiateViewControllerWithIdentifier(storyboardId) as! VC

    return controller
}

这不是类方法。这只是一个函数。这里不需要上课。

let tvc: UITableViewController = instantiateViewController(name: name, storyboardId: storyboardId)

Answer 3

更清洁的解决方案（至少视觉上更整洁）：

斯威夫特 5.1

class func initialize(storyboardName: String, storyboardId: String) -> Self {
    UIStoryboard(name: storyboardName, bundle: nil)
        .instantiateViewController(withIdentifier: storyboardId).view as! Self
}

Answer 4

另一种方法是使用协议，它也允许您返回

Self

。

protocol StoryboardGeneratable {

}

extension UIViewController: StoryboardGeneratable {

}

extension StoryboardGeneratable where Self: UIViewController
{
    static func initialize(storyboardName: String, storyboardId: String) -> Self
    {
        let storyboad = UIStoryboard(name: storyboardName, bundle: nil)
        let controller = storyboad.instantiateViewController(withIdentifier: storyboardId) as! Self
        return controller
    }
}