super giving the wrong type when subclassing tuple()

Question

I tried porting something from python2 to python3, and was confronted with an error concerning the following class:

class Bound(tuple):
    # some methods skipped…

    def __new__(cls, value, is_closed):
        if value is NegativeInfinity or value is PositiveInfinity:
            is_closed = False
        return tuple.__new__(cls, (value, is_closed))

    def __init__(self, value, is_closed):
        """
        See __new__
        """
        super(Bound, self).__init__((value, is_closed))

When trying to initialize, it would fail with object.__init__() takes no parameters. It appears that super(Bound, self).__init__(…) accesses the __init__ method of object, which seems wrong – doesn't super just advance at the objects __mro__?

To narrow things down, I wrote the following construction:

class T(tuple):
    def __new__(cls, arg):
        return super(T, cls).__new__(cls, arg)

    def __init__(self, arg):
        return super(T, self).__init__(arg)

In this example, I get the same error: T([]) tells me object.__init__() takes no parameters nonetheless.

Since T.__mro__ is (__main__.T, tuple, object), this is inherently confusing. The exact same happens when using super() without explicitly stating type and instance.

What goes wrong?

Answer 1

It appears that super(Bound, self).__init__(…) accesses the __init__ method of object, which seems wrong – doesn't super just advance at the objects __mro__?

object.__init__ is the next __init__ in the MRO, because tuple doesn't have its own __init__.

On Python 2, this would have given you a (suppressed by default) warning. On Python 3, it's an error. You can either remove your __init__ method (which, due to special handling in object.__init__, will stop object.__init__ from complaining), or you can call super(Bound, self).__init__(), without arguments.

For more details, see the explanatory comment in the Python source code.