python - Assuming `obj` has type `objtype`, are `super(cls,obj)` and `super(cls,objtype)` the same? -


in python, assuming obj has type objtype, super(cls,obj) , super(cls,objtype) same?

is correct super(cls,obj) converts obj object class superclass of objtype after cls in mro of objtype?

what super(cls,objtype) mean then?

for example, given implementation of singleton design pattern:

class singleton(object):     _singletons = {}     def __new__(cls, *args, **kwds):         if cls not in cls._singletons:             cls._singletons[cls] = super(singleton, cls).__new__(cls)         return cls._singletons[cls]

any subclass of singleton (that not further override __new__ ) has 1 instance.

what super(singleton, cls) mean, cls class? return?

thanks.

according docs, super

return proxy object delegates method calls parent or sibling class of type.

so super returns object knows how call methods of other classes in class hierarchy.

the second argument super object super bound; instance of class, if super being called in context of method classmethod or staticmethod want call method on class object rather instance.

so calling super(someclass, cls).some_method() means call some_method on classes someclass descends from, rather on instances of these classes. otherwise super calls behave super call in instance method.

the usage looks more natural in less complicated code:

class c(somebaseclass):      def method(self):         # bind super 'self'         super(c, self).method()      @classmethod     def cmethod(cls):         # no 'self' here -- bind cls         super(c, cls).cmethod()

note super(c, cls) required in python2, empty super() enough in python3.

in singleton example, super(singleton, cls).__new__(cls) returns result of calling object.__new__(cls), instance of singleton. it's being created way avoid recursively calling singleton.__new__.


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