swift - 如何在当前类型 Swift 中使用泛型函数

Question

我有一个带有通用函数的协议(protocol)，它返回通用 ViewController。它迫使我在默认实现中添加类型约束检查。

protocol RouterViewController {
    func getViewController<T: UIViewController>(_ name: Storyboard) -> T
}

extension RouterViewController {
    func getViewController<T>(_ name: Storyboard) -> T where T: UIViewController {
        let storyBoard = UIStoryboard.init(storyboard: name)
        return storyBoard.instantiateViewController()
    }
}

我可以在没有泛型的情况下完成相同的示例

    struct RouterViewController {
     static func  getViewController(_ name: Storyboard) -> UIViewController {
     let storyBoard = UIStoryboard.init(storyboard: name)
            return storyBoard.instantiateViewController()
 }
}

我有一系列问题

1. 如果我必须进行类型检查，那么我会选择工厂方法吗？如何对泛型做同样的事情？
2. 如果我想使用泛型，如何检查此函数返回中的当前类型，如 func<SignIn>() -> T

Answer 1

如果您对协议(protocol)声明/扩展进行一些小的更改，您将获得所需的功能:

protocol RouterViewController {
    func getViewController<T: UIViewController>(_ viewControllerType: T.Type, _ name: Storyboard) -> T 
}

extension RouterViewController {
    func getViewController<T>(_ viewControllerType: T.Type, _ name: Storyboard) -> T where T: UIViewController {
        let storyBoard = UIStoryboard.init(storyboard: name)
        let identifier = String(describing: T.self) // Let's say view controller's storyboard identifier is same as class name
        return storyBoard.instantiateViewController(withIdentifier: identifier) as! T
    }
}

用法:

struct TestStruct: RouterViewController {

}

TestStruct().getViewController(TestViewController.self, #YourStoryboardObject)

有关 View Controller 和 Storyboard的更多信息，请检查:Swift protocol extension