在Swift中返回实例类型

在Swift中返回实例类型

与在Swift的类扩展函数中使用’self’类似，您可以定义一个通用的辅助方法，该方法可以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")

进行编译，并将类型推断为

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)    }}