super

Most overloaded methods require [super theMethod] call. (For example, [super viewDidLoad]; , [super viewWillAppear]; and [super dealloc]; ) I didn't think twice whether I need [super touchesBegan:withEvent:] call or not, but it seems to play a role somehow. When do I need it and when don't I need it? I'm trying to programmatically cancel touch events when I need to, and it seems to be related to the question I asked. MHC While it all depends on the expected behaviors of the object, let's point out the basic: You call the superclass implementation in an overridden method (you mean overriding,

There is two classes Super1 and Sub1 Super1.class public class Super1 { Super1 (){ this.printThree(); } public void printThree(){ System.out.println("Print Three"); } } Sub1.class public class Sub1 extends Super1 { Sub1 (){ super.printThree(); } int three=(int) Math.PI; public void printThree(){ System.out.println(three); } public static void main(String ...a){ new Sub1().printThree(); } } When I invoke the method printThree of class Sub1 I expected the output to be: Print Three 3 Because Sub1 constructor calling the super.printThree(); . But I actually get 0 Print Three 3 I know 0 is default

class Works(type): def __new__(cls, *args, **kwargs): print([cls,args]) # outputs [<class '__main__.Works'>, ()] return super().__new__(cls, args) class DoesNotWork(type): def __new__(*args, **kwargs): print([args[0],args[:0]]) # outputs [<class '__main__.doesNotWork'>, ()] return super().__new__(args[0], args[:0]) Works() # is fine DoesNotWork() # gets "RuntimeError: super(): no arguments" As far as I can see, in both cases super._new__ receives the class literal as first argument, and an empty tuple as the 2nd. So why does one give an error and the other not? The zero-argument form of super