在 Swift 中返回实例类型

Question

我正在尝试做这个扩展：

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“自我”

有可能吗？我也想让它成为init(storyboardName: String, storyboardId: String)

Answer 1

类似于Using 'self' in class extension functions in Swift，你可以定义一个通用的辅助方法，从调用上下文中推断出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。

---

Swift 3 更新：

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，而不是在运行时确定的。在您的代码中，Self 完全等同于 UIViewController，而不是“碰巧调用它的子类”。这将返回UIViewController，而调用者必须将as 放入正确的子类中。我认为这就是您要避免的（尽管这是“正常的 Cocoa”方式，所以只返回 UIViewController 可能是最好的解决方案）。

注意：在任何情况下都不应将函数命名为 initialize。这是NSObject 的现有类函数，充其量会引起混乱，最坏的情况是会导致错误。

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。

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。这只是一个功能。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（至少在视觉上更整洁）：

斯威夫特 5.1

class func initialize(storyboardName: String, storyboardId: String) -> Self {
    return 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
    }
}