Issue
I have a metaclass:
class MyMeta(type):
def __init__(cls, name, bases, dct):
# Do something
...
return super(MyMeta, cls).__init__(cls, name, bases, dct)
and a class:
class MyClass(object):
__metaclass__ = MyMeta
When I use these I get the following error:
TypeError: Error when calling the metaclass bases
type.__init__() takes 1 or 3 arguments
What's the problem, and why does type.__init__()
take a precisely variable number of arguments?
Solution
The problem is that in the upgrade from python 2.5 to python 2.6 type.__init__()
was changed so that you are no longer required to pass in cls
. So simply make the super
call:
return super(MyMeta, cls).__init__(name, bases, dct)
Another solution is to avoid the super
call altogether and do this (although it's a little less nice):
return type.__init__(cls, name, bases, dct)
And everything will work fine (in python >= 2.6).
As to why type.__init__()
can take differing numbers of arguments, check out the documentation. It's so that as well as using it as a constructor, you can call type(myobject)
and it will return the type of myobject
:
>>> number = 1
>>> type(number)
<type 'int'>
>>> type('my string')
<type 'str'>
See What is a metaclass in Python? for more information on metaclasses and type.
Answered By - Harley Holcombe
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