python metaclass inheritance issue

Question

I have a somewhat bizarre metaclass question. I'm using a meta class to dynamically create a 'sibling' class that inherits from another superclass and assign it as an attribute on the original class. Below is a minimal setup:

class Meta(type):
def __new__(cls, name, parents, dct):
    sdct = dct.copy()
    dct['sibling'] = type(name+'Sibling', (Mom,), sdct)
    return super().__new__(cls, name, (Dad,), dct)

class Mom:
     def __init__(self):
         self.x = 3

class Dad:
     def __init__(self):
         self.x = 4

class Child(metaclass=Meta):
     def __init__(self):
         super().__init__()
         self.y = 1  # <<< added from feedback
print(Child().x) # 4
print(Child().sibling) # <class '__main__.Child'> | should be ChildSibling
print(Child().sibling().x) # should be 3 instead throws:
    # TypeError: super(type, obj): obj must be an instance or subtype of type
print(Child().sibling().y) # should print 4

Something seems to be going wrong above with the creation of the 'sibling' class but I'm not quite sure what. I know for example that this would work:

class ChildAbstract:
    def __init__(self):
        super().__init__()

ChildSibling = type('ChildSibling', (ChildAbstract, Mom), {})
Child = type('Child', (ChildAbstract, Dad), {'sibling': ChildSibling})
print(Child().sibling().x) # 3

I can't see the difference between the two cases though.

Ilja Everilä · Accepted Answer · 2016-03-31 07:46:22Z

6

The dictionary sdct passed to type includes __qualname__, which according to this PEP is what repr and str now use.

Try adding

print(Child is Child.sibling)  # False
print(Child.sibling.__name__)  # "ChildSibling"

and you'll see that it is truly the sibling.

As to why sibling().x throws, the very same sdct also already contains Child.__init__, which ends up as __init__ of your dynamically created new type ChildSibling. During a call to sibling() the super() call resolves the class to Child and is given an instance of ChildSibling:

Also note that, aside from the zero argument form, super() is not limited to use inside methods. The two argument form specifies the arguments exactly and makes the appropriate references. The zero argument form only works inside a class definition, as the compiler fills in the necessary details to correctly retrieve the class being defined, as well as accessing the current instance for ordinary methods.

https://docs.python.org/3/library/functions.html#super

Accessing the current instance is done by passing the first argument to method as instance.

super() -> same as super(__class__, <first argument>)

The error is raised at line 7210 of Object/typeobject.c.

Try removing the wrong __init__ in your __new__ with:

del sdct['__init__']

and now

print(Child().sibling().x)

will print 3.

A solution to "generic" inheritance and meta programming friendlier __init__ is to use the 2 argument form of super():

def __init__(self):
    super(self.__class__, self).__init__()

edited Mar 31, 2016 at 7:46

answered Mar 30, 2016 at 21:22

Ilja Everilä

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5 Comments

Buck Over a year ago

Interesting point, deleting init from the dictionary works but only because in my toy example I don't do anything in Child.__init__ so this doesn't quite solve the problem (updated code to illustrate). I'll take a look at qualname though

Buck Over a year ago

Ahh the note on super explains this

Ilja Everilä Over a year ago

@Buck The reason your second example works is that the bases of those 2 dynamic types include ChildAbstract and so super() is happy.

Buck Over a year ago

so super().__init__() -> super(self.__class__, self).__init__() sovles this, thanks @Ilja

Ilja Everilä Over a year ago

@Buck included your solution as part of the answer too, so it's easier to find

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