Imagine you have a sequence of views in your storyboard:
A -> ... -> Z
You want to have a button on view Z which allows the user to go all the way back to A.
So what you need to do is give the view that you want to back all the way out to, in this case, A, an instance method which is marked as a IBAction and takes in a single parameter of type UIStoryboardSegue *. The name of the method and variable don't matter. What you do within the implementation doesn't matter, either. Here's an example:
Obj-C:
In A's Interface (not Z's):
- (IBAction)cancelSignup:(UIStoryboardSegue *)unwindSegue;
In A's Implementation (not Z's):
- (IBAction)cancelSignup:(UIStoryboardSegue *)unwindSegue {
// Only "implemented" to satisfy a respondsToSelector: search.
// You can actually implement more stuff here, if you want, IE, if
// you need to reach out to a server to mention that this screen was
// returned to from a later screen.
}
Swift:
In A's source (not Z's):
@IBAction func cancelSignup(unwindSegue: UIStoryboardSegue) {
// Only "implemented" to satisfy a respondsToSelector: search.
// You can actually implement more stuff here, if you want, IE, if
// you need to reach out to a server to mention that this screen was
// returned to from a later screen.
}
Now, within your storyboard, control drag from an element on Z (IE, a cancel button) to Z's Exit. It'll scan through all of the views higher up in the view hierarchy which have an IBAction that accepts only a single UIStoryboardSegue * as an action and list them for you to pick from.
Hopefully this was more straight forward and helpful than the existing answers. I found that this link was particularly useful, so if there's a detail you're still fuzzy on after reading my answer, maybe this can help you (I tried to just condense all the useful info from this long article into a short answer):
http://www.freelancemadscience.com/fmslabs_blog/2012/9/24/advanced-storyboard-techniques.html