Using abc.ABCMeta in a way it is compatible both with Python 2.7 and Python 3.5

Question

I'd like to create a class which has abc.ABCMeta as a metaclass and is compatible both with Python 2.7 and Python 3.5. Until now, I only succeeded doing this either on 2.7 or on 3.5 - but never on both versions simultaneously. Could someone give me a hand?

Python 2.7:

import abc
class SomeAbstractClass(object):
    __metaclass__ = abc.ABCMeta
    @abc.abstractmethod
    def do_something(self):
        pass

Python 3.5:

import abc
class SomeAbstractClass(metaclass=abc.ABCMeta):
    @abc.abstractmethod
    def do_something(self):
        pass

Testing

If we run the following test using the suitable version of the Python interpreter (Python 2.7 -> Example 1, Python 3.5 -> Example 2), it succeeds in both scenarios:

import unittest
class SomeAbstractClassTestCase(unittest.TestCase):
    def test_do_something_raises_exception(self):
        with self.assertRaises(TypeError) as error:
            processor = SomeAbstractClass()
        msg = str(error.exception)
        expected_msg = "Can't instantiate abstract class SomeAbstractClass with abstract methods do_something"
        self.assertEqual(msg, expected_msg)

Problem

While running the test using Python 3.5, the expected behavior doesn't happen (TypeError is not raised while instantiating SomeAbstractClass):

======================================================================
FAIL: test_do_something_raises_exception (__main__.SomeAbstractClassTestCase)
----------------------------------------------------------------------
Traceback (most recent call last):
  File "/home/tati/sample_abc.py", line 22, in test_do_something_raises_exception
    processor = SomeAbstractClass()
AssertionError: TypeError not raised

----------------------------------------------------------------------

Whereas running the test using Python 2.7 raises a SyntaxError:

 Python 2.7 incompatible
 Raises exception:
  File "/home/tati/sample_abc.py", line 24
    class SomeAbstractClass(metaclass=abc.ABCMeta):
                                     ^
 SyntaxError: invalid syntax

Answer 1

You could use six.add_metaclass or six.with_metaclass:

import abc, six

@six.add_metaclass(abc.ABCMeta)
class SomeAbstractClass():
    @abc.abstractmethod
    def do_something(self):
        pass

six is a Python 2 and 3 compatibility library. You can install it by running pip install six or by downloading the latest version of six.py to your project directory.

For those of you who prefer future over six, the relevant function is future.utils.with_metaclass.