Return Swift 中的实例类型

Return instancetype in Swift

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:

error: cannot convert return expression of type 'UIViewController' to return type 'Self'

可能吗?我也想把它变成 init(storyboardName: String, storyboardId: String)

Self 是在编译时确定的,而不是运行时。在您的代码中,Self 完全等同于 UIViewController,而不是 "the subclass that happens to be calling this." 这将转到 return UIViewController 并且调用者必须 as它进入右子class。我认为这就是您要避免的(尽管这是 "normal Cocoa" 的方法,所以 returning 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 class 函数,最好的情况下会造成混乱,最坏的情况下会导致错误。

但是如果你想避免调用者的 as,subclassing 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,泛型就是您所需要的。

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 class 方法。这只是一个功能。这里不需要 class。

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

类似,您可以定义一个通用的辅助方法,它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.


Swift3 的更新:

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)
    }
}

另一种方法是使用协议,它还允许您 return 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
    }
}

更简洁的解决方案(至少在视觉上更整洁):

Swift 5.1

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