Chain-calling parent initialisers in python
Question:
Consider this – a base class A, class B inheriting from A, class C inheriting from B. What is a generic way to call a parent class initialiser in an initialiser? If this still sounds too vague, here’s some code.
class A(object):
def __init__(self):
print "Initialiser A was called"
class B(A):
def __init__(self):
super(B,self).__init__()
print "Initialiser B was called"
class C(B):
def __init__(self):
super(C,self).__init__()
print "Initialiser C was called"
c = C()
This is how I do it now. But it still seems a bit too non-generic – you still must pass a correct type by hand.
Now, I’ve tried using self.__class__
as a first argument to super(), but, obviously it doesn’t work – if you put it in the initialiser for C – fair enough, B’s initialiser gets called. If you do the same in B, “self” still points to an instance of C so you end up calling B’s initialiser again (this ends in an infinite recursion).
There is no need to think about diamond inheritance for now, I am just interested in solving this specific problem.
Answers:
The way you are doing it is indeed the recommended one (for Python 2.x).
The issue of whether the class is passed explicitly to super
is a matter of style rather than functionality. Passing the class to super
fits in with Python’s philosophy of “explicit is better than implicit”.
Python 3 includes an improved super() which allows use like this:
super().__init__(args)
You can simply write :
class A(object):
def __init__(self):
print "Initialiser A was called"
class B(A):
def __init__(self):
A.__init__(self)
# A.__init__(self,<parameters>) if you want to call with parameters
print "Initialiser B was called"
class C(B):
def __init__(self):
# A.__init__(self) # if you want to call most super class...
B.__init__(self)
print "Initialiser C was called"
Consider this – a base class A, class B inheriting from A, class C inheriting from B. What is a generic way to call a parent class initialiser in an initialiser? If this still sounds too vague, here’s some code.
class A(object):
def __init__(self):
print "Initialiser A was called"
class B(A):
def __init__(self):
super(B,self).__init__()
print "Initialiser B was called"
class C(B):
def __init__(self):
super(C,self).__init__()
print "Initialiser C was called"
c = C()
This is how I do it now. But it still seems a bit too non-generic – you still must pass a correct type by hand.
Now, I’ve tried using self.__class__
as a first argument to super(), but, obviously it doesn’t work – if you put it in the initialiser for C – fair enough, B’s initialiser gets called. If you do the same in B, “self” still points to an instance of C so you end up calling B’s initialiser again (this ends in an infinite recursion).
There is no need to think about diamond inheritance for now, I am just interested in solving this specific problem.
The way you are doing it is indeed the recommended one (for Python 2.x).
The issue of whether the class is passed explicitly to super
is a matter of style rather than functionality. Passing the class to super
fits in with Python’s philosophy of “explicit is better than implicit”.
Python 3 includes an improved super() which allows use like this:
super().__init__(args)
You can simply write :
class A(object):
def __init__(self):
print "Initialiser A was called"
class B(A):
def __init__(self):
A.__init__(self)
# A.__init__(self,<parameters>) if you want to call with parameters
print "Initialiser B was called"
class C(B):
def __init__(self):
# A.__init__(self) # if you want to call most super class...
B.__init__(self)
print "Initialiser C was called"