Why does `if None.__eq__("a")` seem to evaluate to True (but not quite)?
As you already figured None.__eq__("a")
evaluates to NotImplemented
however if you try something like
if NotImplemented:
print("Yes")
else:
print("No")
the result is
yes
this mean that the truth value of NotImplemented
true
Therefor the outcome of the question is obvious:
None.__eq__(something)
yields NotImplemented
And bool(NotImplemented)
evaluates to True
So if None.__eq__("a")
is always True
The result you are seeing is caused by that fact that
None.__eq__("a") # evaluates to NotImplemented
evaluates to NotImplemented
, and NotImplemented
's truth value is documented to be True
:
https://docs.python.org/3/library/constants.html
Special value which should be returned by the binary special methods (e.g.
__eq__()
,__lt__()
,__add__()
,__rsub__()
, etc.) to indicate that the operation is not implemented with respect to the other type; may be returned by the in-place binary special methods (e.g.__imul__()
,__iand__()
, etc.) for the same purpose. Its truth value is true.
If you call the __eq()__
method manually rather than just using ==
, you need to be prepared to deal with the possibility it may return NotImplemented
and that its truth value is true.
This is a great example of why the __dunder__
methods should not be used directly as they are quite often not appropriate replacements for their equivalent operators; you should use the ==
operator instead for equality comparisons, or in this special case, when checking for None
, use is
(skip to the bottom of the answer for more information).
You've done
None.__eq__('a')
# NotImplemented
Which returns NotImplemented
since the types being compared are different. Consider another example where two objects with different types are being compared in this fashion, such as 1
and 'a'
. Doing (1).__eq__('a')
is also not correct, and will return NotImplemented
. The right way to compare these two values for equality would be
1 == 'a'
# False
What happens here is
- First,
(1).__eq__('a')
is tried, which returnsNotImplemented
. This indicates that the operation is not supported, so 'a'.__eq__(1)
is called, which also returns the sameNotImplemented
. So,- The objects are treated as if they are not the same, and
False
is returned.
Here's a nice little MCVE using some custom classes to illustrate how this happens:
class A:
def __eq__(self, other):
print('A.__eq__')
return NotImplemented
class B:
def __eq__(self, other):
print('B.__eq__')
return NotImplemented
class C:
def __eq__(self, other):
print('C.__eq__')
return True
a = A()
b = B()
c = C()
print(a == b)
# A.__eq__
# B.__eq__
# False
print(a == c)
# A.__eq__
# C.__eq__
# True
print(c == a)
# C.__eq__
# True
Of course, that doesn't explain why the operation returns true. This is because NotImplemented
is actually a truthy value:
bool(None.__eq__("a"))
# True
Same as,
bool(NotImplemented)
# True
For more information on what values are considered truthy and falsy, see the docs section on Truth Value Testing, as well as this answer. It is worth noting here that NotImplemented
is truthy, but it would have been a different story had the class defined a __bool__
or __len__
method that returned False
or 0
respectively.
If you want the functional equivalent of the ==
operator, use operator.eq
:
import operator
operator.eq(1, 'a')
# False
However, as mentioned earlier, for this specific scenario, where you are checking for None
, use is
:
var = 'a'
var is None
# False
var2 = None
var2 is None
# True
The functional equivalent of this is using operator.is_
:
operator.is_(var2, None)
# True
None
is a special object, and only 1 version exists in memory at any point of time. IOW, it is the sole singleton of the NoneType
class (but the same object may have any number of references). The PEP8 guidelines make this explicit:
Comparisons to singletons like
None
should always be done withis
oris not
, never the equality operators.
In summary, for singletons like None
, a reference check with is
is more appropriate, although both ==
and is
will work just fine.