Question

我试图进行此扩展：

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
    }
}

但我收到编译错误：

错误：无法转换类型＆＃39; UIViewController＆＃39;的返回表达式至返回类型＆＃39;自我＆＃39;

有可能吗？我也希望将其设为init(storyboardName: String, storyboardId: String)

Answer 1

与Using 'self' in class extension functions in Swift类似，您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。

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在编译时确定，而不是运行时。在您的代码中，UIViewController完全等同于UIViewController，而不是＆＃34;正好调用此类的子类。＆＃34;这将返回as，调用者必须UIViewController进入正确的子类。我认为这是你想要避免的（尽管它是＃34;正常的Cocoa＆＃34;这样做的方式，所以只返回initialize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NSObject。这是as的现有类函数，最多会造成混淆，最坏的情况是错误。

但是如果你想避免调用者的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 }，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，只需要泛型。

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

这不是一种类方法。它只是一个功能。这里没有必要上课。

numberOfSectionsInTableView

Answer 3

另一种方法是使用协议，该协议还允许您返回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
    }
}

Answer 4

更清洁的解决方案（至少在视觉上更整洁）：

class func initialize(storyboardName: String, storyboardId: String) -> Self {
    // The absurdity that's Swift's type system. If something is possible to do with two functions, why not let it be just one?
    func loadFromImpl<T>() -> T {
        let storyboard = UIStoryboard(name: storyboardName, bundle: nil)
        return storyboard.instantiateViewController(withIdentifier: storyboardId).view as! T
    }
    return loadFromImpl()
}

在Swift中返回instancetype

4 个答案: