If you execute the following statement in Python 3.7, it will (from my testing) print
if None.__eq__("a"): print("b")
None.__eq__("a") evaluates to
"a".__eq__("a") evaluates to
"b".__eq__("a") evaluates to
I initially discovered this when testing the return value of a function, but didn't return anything in the second case -- so, the function returned
What's going on here?
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
is (skip to the bottom of the answer for more information).
None.__eq__('a') # NotImplemented
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).__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
(1).__eq__('a')is tried, which returns
NotImplemented. This indicates that the operation is not supported, so
'a'.__eq__(1)is called, which also returns the same
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
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
__len__ method that returned
If you want the functional equivalent of the
== operator, use
import operator operator.eq(1, 'a') # False
However, as mentioned earlier, for this specific scenario, where you are checking for
var = 'a' var is None # False var2 = None var2 is None # True
The functional equivalent of this is using
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
Noneshould always be done with
is not, never the equality operators.
In summary, for singletons like
None, a reference check with
is is more appropriate, although both
is will work just fine.
The result you are seeing is caused by that fact that
None.__eq__("a") # evaluates to NotImplemented
NotImplemented's truth value is documented to be
Special value which should be returned by the binary special methods (e.g.
__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.
__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.
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
this mean that the truth value of
Therefor the outcome of the question is obvious:
bool(NotImplemented) evaluates to True
if None.__eq__("a") is always True