在Swift中返回instancetype

发布时间：2020-12-14 05:58:52 所属栏目：百科来源：网络整理

导读：我试图做这个扩展： extension UIViewController{ class func initialize(storyboardName: String,storyboardId: String) - Self { let storyboad = UIStoryboard(name: storyboardName,bundle: nil) let controller = storyboad.instantiateViewControllerW

我试图做这个扩展：

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：

error: cannot convert return expression of type ‘UIViewController’ to
return type ‘Self’

可能吗？我也想做为init(storyboardName：String，storyboardId：String)

类似于 Using ‘self’ in class extension functions in Swift，您可以定义一个通用辅助方法，从调用上下文中推断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的更新：

extension UIViewController
{
    class func instantiateFromStoryboard(storyboardName: String,storyboardId: String) -> Self
    {
        return instantiateFromStoryboardHelper(storyboardName: storyboardName,bundle: nil)
        let controller = storyboard.instantiateViewController(withIdentifier: storyboardId) as! T
        return controller
    }
}

（编辑：李大同）

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