在 Swift 中返回实例类型 [英] Return instancetype in Swift

问题描述

我正在尝试进行此扩展:

I'm trying to make this extension:

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自我"

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

有可能吗?我也想把它设为 init(storyboardName: String, storyboardId: String)

Is it possible? Also I want to make it as init(storyboardName: String, storyboardId: String)

推荐答案

类似于 在 Swift 的类扩展函数中使用 'self'，您可以定义一个通用的辅助方法，该方法从调用上下文推断 self 的类型:

Similar as in Using 'self' in class extension functions in Swift, you can define a generic helper method which infers the type of self from the calling context:

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 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:

Another possible solution, using 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)
    }
}

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