By asmeurer


2014-06-27 15:23:36 8 Comments

According to the Python docs: "when defining __eq__(), one should also define __ne__() so that the operators will behave as expected".

However, it appears that Python computes __ne__ as not __eq__ automatically:

In [8]: class Test:
    def __eq__(self, other):
        print("calling __eq__")
   ...:         return isinstance(other, Test)
   ...:

In [9]: a = Test()

In [10]: b = Test()

In [11]: a == b
calling __eq__
Out[11]: True

In [12]: a != b
calling __eq__
Out[12]: False

In [13]: a == 1
calling __eq__
Out[13]: False

In [14]: a != 1
calling __eq__
Out[14]: True

So what's the point of defining __ne__ if it's just going to be return not self.__eq__(other)? And furthermore, where is this behavior actually documented?

EDIT

Apparently it matters that I am using Python 3. In Python 2, I get

In [1]: class Test(object):
   ...:     def __eq__(self, other):
   ...:         print("calling __eq__")
   ...:         return isinstance(other, Test)
   ...:

In [2]: a = Test()

In [3]: b = Test()

In [4]: a == b
calling __eq__
Out[4]: True

In [5]: a != b
Out[5]: True

In [6]: a == 1
calling __eq__
Out[6]: False

In [7]: a != 1
Out[7]: True

But the docs I referenced are the Python 3 docs. Were they just not updated?

1 comments

@Martijn Pieters 2014-06-27 15:55:30

Python 3 changed behaviour for the == case, see Python 3, What's New:

!= now returns the opposite of ==, unless == returns NotImplemented.

It was deemed a useful change.

The fact that the documentation has not been updated is indeed a long standing bug.

However, as a comment on the report points out, if you inherit from a class that already has defined __ne__, overriding just __eq__ is not enough and you'll also have to override the __ne__ method.

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