ios - Unwind Segue Programmatically - New twist -

context: have 2 vc, , b. vc contains several buttons , several labels. when pressing button in vc a segue display vc b/c/ , forth. now, when finished vc a/b/c forth, segue being unwind vc appears. of vc b/c/d forth, using unwind method trigger through button in vc (ctrl + drag "exit" icon). works perfect, because upon returning vc a, following action being called automatically:

- (ibaction)returned:(uistoryboardsegue *)segue { // here stuff  } 

problem: now, in 1 of secondary vcs (e.g. d), things bit special. in vc generate hundred buttons through loop programmatically, detect button being pressed , unwind vc (without specific button; of buttons trigger unwind). know can eg using this

[self dismissviewcontrolleranimated: yes completion: nil] 

but not trigger above action when returning vc a, or using this

[self performseguewithidentifier:@"unwindsegueidentifier" sender:self] 

but generate new instance of vc a, not want (because labels in instance of vc contains information).

so, want able return same instance of vc generated vc d, , trigger "returned" action listed above. thus, want achieve same effect when using button connected "exit" icon, want programmatically "inside code" when 1 many buttons in vc d pressed.

any thoughts?

[self performseguewithidentifier:@"unwindsegueidentifier" sender:self] should work fine. unwind segues not create new instances of destinations, no matter how they're performed. unique in it's segue not create new instance of destination.