Question

我正在尝试制作一个函数，该函数接收View Controller的名称（字符串），然后以模态形式显示该View Controller。示例：

func presentViewControllerModally(newViewControllerName: String){

     let storyBoard: UIStoryboard = UIStoryboard(name: "Main", bundle: nil)
     let newViewController = storyBoard.instantiateViewController(withIdentifier: "\(newViewControllerName)_Identifier") as! newViewControllerName
     self.present(newViewController, animated: false, completion: nil)
}

但是我无法使as! newViewControllerName像newViewControllerName的类型一样工作，而不是对象本身。

Answer 1

您不必将控制器转换为自定义子类。您可以只显示UIViewController。

Answer 2

如果必须返回真正的viewController类型，只需使用泛型即可。

R *r;

void on_recv(struct bufferevent *bev, void *arg)
{
    struct evbuffer *src;
    size_t len;

    src = bufferevent_get_input(bev);
    len = evbuffer_get_length(src);
    ......

    BaseHandler *handler = r->get_handler();
    char data[MAX_BUFSIZE] = { 0 };
    char new_data[MAX_BUFSIZE] = { 0 };
    evbuffer_copyout(src, data, len);
    LOGE("recv len1: %d\n", len);
    if (handler->handle(data, new_data)) {
        LOGE("recv len2: %d\n", len);
    }

    ......

}

然后打电话

func presentViewControllerModally<T: UIViewController>(newViewControllerName: String) -> T? {
    let storyBoard: UIStoryboard = UIStoryboard(name: "Main", bundle: nil)
    if let newViewController = storyBoard.instantiateViewController(withIdentifier: "\(newViewControllerName)_Identifier") as? T {
        self.present(newViewController, animated: false, completion: nil)
        return newViewController
    }
    return nil
}

如果您不需要正确的类型，而只是UIViewController，则不需要强制转换。

是否可以使编译器将VC名称（从字符串中）视为该VC类型？

2 个答案: