ios - 无法传递 UIVewController 的实例 - iOS

Question

我有一个带有我制作的方法的应用程序，想法是这个方法将打开我想要的任何 View Controller 。我做到了，所以我必须拥有许多相同代码的副本。但是，它不起作用，Xcode 不断出现错误:

Invalid operands to binary expression (UIViewController ) and 'unsigned long()(const char*)')

这是我的方法:

-(void)open_view_controller:(UIViewController *)controller_name :(NSString *)view_name {

    controller_name *screen = [[controller_name alloc] initWithNibName:view_name bundle:nil];
    screen.modalTransitionStyle = UIModalTransitionStyleCoverVertical;
    [self presentViewController:screen animated:YES completion:nil];
}

我只是想避免这样做:

CallCreator *screen = [[CallCreator alloc] initWithNibName:@"CallCreator" bundle:nil];
screen.modalTransitionStyle = UIModalTransitionStyleCoverVertical;
[self presentViewController:screen animated:YES completion:nil];

我能做些什么来解决这个错误？我不明白出了什么问题，我传入了正确的类型 - UIViewController .

谢谢你的时间，丹。

Answer 1

您应该使用 Class而不是 UIViewController *

- (void)presentViewControllerOfClass:(Class)controllerClass viewName:(NSString *)viewName;
{
    UIViewController *screen = [[controllerClass alloc] initWithNibName:viewName bundle:nil];
    screen.modalTransitionStyle = UIModalTransitionStyleCoverVertical;
    [self presentViewController:screen animated:YES completion:nil];
}

这将像

[self presentViewControllerOfClass:MyViewController.class viewName:@"MyViewController"];