In my Iphone app I give a button when pressed calls a specific number using the following function call and uses the native iphone call app.
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:call_url]];
where call_url is of the form tel://
Once the call is done, is there a way to restore and open my Iphone App as it was before calling ?
Best Answer
No. Your process has been terminated. You'll have to depend on the user to start your app again and you have to restore the state yourself.