UIStoryboard: What's the Correct Way to Get the Active Storyboard?

In case you want to get the active storyboard for a viewController, there's a storyboard property. This is how I solved it, instead of making a new instance:

LoginViewController *vc = [navController.storyboard instantiateViewControllerWithIdentifier:@"firstLaunch"];
[navController presentModalViewController:vc animated:YES];

In Swift you'd call:

let loginViewController = navigationController?.storyboard?.instantiateViewController(withIdentifier: "firstLaunch") as! LoginViewController
navigationController?.present(loginViewController, animated: true, completion: nil)

You could also be a lot safer by using guards against the navigation controller and the storyboard. I've used as! so as to guarantee that you're getting a LoginController.

OK. As my comment above indicates, I found the answer to the (badly phrased question):

I wanted to be able to get the main (not active) storyboard, as I'm not using multiple storyboards per incarnation. I'm using the standard model of 1 storyboard for iPhone, and 1 for iPad. I just wanted the cleanest way to get the storyboard, so that I could use it to generate a view controller.

I found the answer in this post on Stack Overflow, and implemented it with the following code:

UIStoryboard *st = [UIStoryboard storyboardWithName:[[NSBundle mainBundle].infoDictionary objectForKey:@"UIMainStoryboardFile"] bundle:[NSBundle mainBundle]];

In Swift, you'd use the following syntax:

let storyboard = UIStoryboard(name: "Main", bundle: nil) 

Note that passing nil to bundle will make the call refer to your main bundle automatically.

If you're in a view controller that you have on the Storyboard and want to instantiate the Storyboard from there directly, you can just do:

let storyboard: UIStoryboard? = self.storyboard // call this inside a VC that is on the Storyboard

Note that in the last case, self.storyboard will return an optional Storyboard (Storyboard?), so if you'd like to use it unwrap it like so:

if let storyboard = self.storyboard {
  // access storyboard here
}