super

I was wondering, when creating new Activity classes and then overriding the onCreate() method, in eclipse I always get auto added: super.onCreate() . How does this happen? Is there a java keyword in the abstract or parent class that forces this? I don't know if it is illegal not to call the super class, but I remember in some methods that I got a exception thrown for not doing this. Is this also built-in into java? Can you use some keyword to do that? Or how is it done? Here's the source of Activity#onCreate() - it is almost all comments ( original - see line ~800 ): /** * Called when the

In many of Android methods, especially constructors and overridden methods, you should or even must call the parent class method using super() . When you use the Eclipse Source > Override/Implement Methods... you get code from a template with TODO tags like this: public MyCanvas(Context context, AttributeSet attrs) { super(context, attrs); // TODO Auto-generated constructor stub } @Override protected void onDraw(Canvas canvas) { // TODO Auto-generated method stub super.onDraw(canvas); } I do not understand exacly what the superclass does in each case so I always insert my code at the exact

Is it possible to call a super static method from child static method? I mean, in a generic way, so far now I have the following: public class BaseController extends Controller { static void init() { //init stuff } } public class ChildController extends BaseController { static void init() { BaseController.loadState(); // more init stuff } } and it works, but I'd like to do it in a generic way, something like calling super.loadState(), which doesn't seem to work... sgokhales In Java, static methods cannot be overidden. The reason is neatly explained here So, it doesn't depend on the object that

I am working with some code that has 3 levels of class inheritance. From the lowest level derived class, what is the syntax for calling a method 2 levels up the hierarchy, e.g. a super.super call? The "middle" class does not implement the method I need to call. Well, this is one way of doing it: class Grandparent(object): def my_method(self): print "Grandparent" class Parent(Grandparent): def my_method(self): print "Parent" class Child(Parent): def my_method(self): print "Hello Grandparent" Grandparent.my_method(self) Maybe not what you want, but it's the best python has unless I'm mistaken.

I wonder when to use what flavour of Python 3 super (). Help on class super in module builtins: class super(object) | super() -> same as super(__class__, <first argument>) | super(type) -> unbound super object | super(type, obj) -> bound super object; requires isinstance(obj, type) | super(type, type2) -> bound super object; requires issubclass(type2, type) Until now I've used super() only without arguments and it worked as expected (by a Java developer). Questions: What does "bound" mean in this context? What is the difference between bound and unbound super object? When to use super(type,

I'd like to override a method in an Objective C class that I don't have the source to. I've looked into it, and it appears that Categories should allow me to do this, but I'd like to use the result of the old method in my new method, using super to get the old methods result. Whenever I try this though, my method gets called, but "super" is nil... Any idea why? I'm doing iPhone development with the XCode 2.2 SDK. I'm definitely working with an instance of a class, and the method of the class is an instance method. @implementation SampleClass (filePathResolver) -(NSString*)