Understanding Python super() with __init__() methods [duplicate]
super() lets you avoid referring to the base class explicitly, which can be nice. But the main advantage comes with multiple inheritance, where all sorts of fun stuff can happen. See the standard docs on super if you haven't already.
Note that the syntax changed in Python 3.0: you can just say
super().__init__() instead of
super(ChildB, self).__init__() which IMO is quite a bit nicer. The standard docs also refer to a guide to using
super() which is quite explanatory.
I'm trying to understand
The reason we use
super is so that child classes that may be using cooperative multiple inheritance will call the correct next parent class function in the Method Resolution Order (MRO).
In Python 3, we can call it like this:
class ChildB(Base): def __init__(self): super().__init__()
In Python 2, we were required to call
super like this with the defining class's name and
self, but we'll avoid this from now on because it's redundant, slower (due to the name lookups), and more verbose (so update your Python if you haven't already!):
Without super, you are limited in your ability to use multiple inheritance because you hard-wire the next parent's call:
Base.__init__(self) # Avoid this.
I further explain below.
"What difference is there actually in this code?:"
class ChildA(Base): def __init__(self): Base.__init__(self)class ChildB(Base): def __init__(self): super().__init__()
The primary difference in this code is that in
ChildB you get a layer of indirection in the
super, which uses the class in which it is defined to determine the next class's
__init__ to look up in the MRO.
I illustrate this difference in an answer at the canonical question, How to use 'super' in Python?, which demonstrates dependency injection and cooperative multiple inheritance.
If Python didn't have
Here's code that's actually closely equivalent to
super (how it's implemented in C, minus some checking and fallback behavior, and translated to Python):
class ChildB(Base): def __init__(self): mro = type(self).mro() check_next = mro.index(ChildB) + 1 # next after *this* class. while check_next < len(mro): next_class = mro[check_next] if '__init__' in next_class.__dict__: next_class.__init__(self) break check_next += 1
Written a little more like native Python:
class ChildB(Base): def __init__(self): mro = type(self).mro() for next_class in mro[mro.index(ChildB) + 1:]: # slice to end if hasattr(next_class, '__init__'): next_class.__init__(self) break
If we didn't have the
super object, we'd have to write this manual code everywhere (or recreate it!) to ensure that we call the proper next method in the Method Resolution Order!
How does super do this in Python 3 without being told explicitly which class and instance from the method it was called from?
It gets the calling stack frame, and finds the class (implicitly stored as a local free variable,
__class__, making the calling function a closure over the class) and the first argument to that function, which should be the instance or class that informs it which Method Resolution Order (MRO) to use.
Since it requires that first argument for the MRO, using
super with static methods is impossible as they do not have access to the MRO of the class from which they are called.
Criticisms of other answers:
super() lets you avoid referring to the base class explicitly, which can be nice. . But the main advantage comes with multiple inheritance, where all sorts of fun stuff can happen. See the standard docs on super if you haven't already.
It's rather hand-wavey and doesn't tell us much, but the point of
super is not to avoid writing the parent class. The point is to ensure that the next method in line in the method resolution order (MRO) is called. This becomes important in multiple inheritance.
I'll explain here.
class Base(object): def __init__(self): print("Base init'ed")class ChildA(Base): def __init__(self): print("ChildA init'ed") Base.__init__(self)class ChildB(Base): def __init__(self): print("ChildB init'ed") super().__init__()
And let's create a dependency that we want to be called after the Child:
class UserDependency(Base): def __init__(self): print("UserDependency init'ed") super().__init__()
ChildB uses super,
ChildA does not:
class UserA(ChildA, UserDependency): def __init__(self): print("UserA init'ed") super().__init__()class UserB(ChildB, UserDependency): def __init__(self): print("UserB init'ed") super().__init__()
UserA does not call the UserDependency method:
'edChildA init'edBase init'ed<__main__.UserA object at 0x0000000003403BA8>UserA()UserA init
UserB does in-fact call UserDependency because
'edChildB init'edUserDependency init'edBase init'ed<__main__.UserB object at 0x0000000003403438>UserB()UserB init
Criticism for another answer
In no circumstance should you do the following, which another answer suggests, as you'll definitely get errors when you subclass ChildB:
super(self.__class__, self).__init__() # DON'T DO THIS! EVER.
(That answer is not clever or particularly interesting, but in spite of direct criticism in the comments and over 17 downvotes, the answerer persisted in suggesting it until a kind editor fixed his problem.)
self.__class__ as a substitute for the class name in
super() will lead to recursion.
super lets us look up the next parent in the MRO (see the first section of this answer) for child classes. If you tell
super we're in the child instance's method, it will then lookup the next method in line (probably this one) resulting in recursion, probably causing a logical failure (in the answerer's example, it does) or a
RuntimeError when the recursion depth is exceeded.
class Polygon(object): def __init__(self, id): self.id = id...class Rectangle(Polygon): def __init__(self, id, width, height): super(self.__class__, self).__init__(id) self.shape = (width, height)...class Square(Rectangle): pass... Square('a', 10, 10)Traceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 3, in __init__TypeError: __init__() missing 2 required positional arguments: 'width' and 'height'
Python 3's new
super() calling method with no arguments fortunately allows us to sidestep this issue.
It's been noted that in Python 3.0+ you can use
to make your call, which is concise and does not require you to reference the parent OR class names explicitly, which can be handy. I just want to add that for Python 2.7 or under, some people implement a name-insensitive behaviour by writing
self.__class__ instead of the class name, i.e.
super(self.__class__, self).__init__() # DON'T DO THIS!
HOWEVER, this breaks calls to
super for any classes that inherit from your class, where
self.__class__ could return a child class. For example:
class Polygon(object): def __init__(self, id): self.id = idclass Rectangle(Polygon): def __init__(self, id, width, height): super(self.__class__, self).__init__(id) self.shape = (width, height)class Square(Rectangle): pass
Here I have a class
Square, which is a sub-class of
Rectangle. Say I don't want to write a separate constructor for
Square because the constructor for
Rectangle is good enough, but for whatever reason I want to implement a Square so I can reimplement some other method.
When I create a
mSquare = Square('a', 10,10), Python calls the constructor for
Rectangle because I haven't given
Square its own constructor. However, in the constructor for
Rectangle, the call
super(self.__class__,self) is going to return the superclass of
mSquare, so it calls the constructor for
Rectangle again. This is how the infinite loop happens, as was mentioned by @S_C. In this case, when I run
super(...).__init__() I am calling the constructor for
Rectangle but since I give it no arguments, I will get an error.