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?
Python 3 changed behaviour for the ==
case, see Python 3, What's New :
!=
now returns the opposite of==
, unless==
returnsNotImplemented
.
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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