super Essentially, it allows you to use the implementations of the current class’ superclass. For the gritty details of the Objective-C runtime: [super message] has the following meaning: When it encounters a method call, the compiler generates a call to one of the functions objc_msgSend, objc_msgSend_stret, objc_msgSendSuper, or objc_msgSendSuper_stret. Messages sent to an object’s superclass … Read more
No you cannot. The super() call needs to know what class the method is part of, to search the base classes for an overridden method. If you pass in self.__class__ (or better still, type(self)) then super() is given the wrong starting point to search for methods, and will end up calling its own method again. … Read more
C# equivalent of your code is class Imagedata : PDFStreamEngine { // C# uses “base” keyword whenever Java uses “super” // so instead of super(…) in Java we should call its C# equivalent (base): public Imagedata() : base(ResourceLoader.loadProperties(“org/apache/pdfbox/resources/PDFTextStripper.properties”, true)) { } // Java methods are virtual by default, when C# methods aren’t. // So we … Read more
Change your code to this and I think it’ll explain things (presumably super is looking at where, say, B is in the __mro__?): class A(object): def __init__(self): print “A init” print self.__class__.__mro__ class B(A): def __init__(self): print “B init” print self.__class__.__mro__ super(B, self).__init__() class C(A): def __init__(self): print “C init” print self.__class__.__mro__ super(C, self).__init__() class … Read more
Consider the following situation: class A(object): def __init__(self): print(‘Running A.__init__’) super(A,self).__init__() class B(A): def __init__(self): print(‘Running B.__init__’) # super(B,self).__init__() A.__init__(self) class C(A): def __init__(self): print(‘Running C.__init__’) super(C,self).__init__() class D(B,C): def __init__(self): print(‘Running D.__init__’) super(D,self).__init__() foo=D() So the classes form a so-called inheritance diamond: A / \ B C \ / D Running the code yields … Read more
That’s not what super() is for. Super basically picks one (or all) of its parents in a specific order. If you only want to call a single parent’s method, do this class ASDF(ASDF1, ASDF2, ASDF3): def __init__(self): ASDF2.__init__(self)
super to bound a named type parameter (e.g. <S super T>) as opposed to a wildcard (e.g. <? super T>) is ILLEGAL simply because even if it’s allowed, it wouldn’t do what you’d hoped it would do, because since Object is the ultimate super of all reference types, and everything is an Object, in effect … Read more
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. What you’re asking sounds anti-pythonic and you’d have to explain why you’re doing … Read more
The easiest example I can think of is: public static <T extends Comparable<? super T>> void sort(List<T> list) { list.sort(null); } taken from the same Collections. This way a Dog can implement Comparable<Animal> and if Animal already implements that, Dog does not have to do anything. EDIT for a real example: After some email ping-pongs, … Read more
The short answer to Am I calling super(type) incorrectly here or is there something I’m missing? is: yes, you’re calling it incorrectly… AND (indeed, because) there is something you’re missing. But don’t feel bad; this is an extremely difficult subject. The documentation notes that If the second argument is omitted, the super object returned is … Read more