Why does `True == False is False` evaluate to False?
PythonPython Problem Overview
I get some rather unexpected behavior on an expression that works with == but not with is:
>>> (True == False) is False
True
>>> True == (False is False)
True
>>> True == False is False
False
>>> id(True)
8978640
>>> id(False)
8978192
>>> id(True == False)
8978192
>>> id(False is False)
8978640
Python Solutions
Solution 1 - Python
Because in fact that's a chained comparison, so
True == False is False
is equivalent to
(True == False) and (False is False)
This can be surprising in this case, but lets you write 1 <= x < 4 unlike in other languages like C.
Solution 2 - Python
From the docs:
> x < y <= z is equivalent to x < y and y <= z, except that y is > evaluated only once (but in both cases z is not evaluated at all when > x < y is found to be false).
In your case True == False is False is equivalent to True == False and False is False as the first condition is False so it short-circuits and return False.
>>> dis.dis(lambda : True == False is False)
1 0 LOAD_GLOBAL 0 (True)
3 LOAD_GLOBAL 1 (False)
6 DUP_TOP
7 ROT_THREE
8 COMPARE_OP 2 (==)
11 JUMP_IF_FALSE_OR_POP 21 <---------this step
14 LOAD_GLOBAL 1 (False)
17 COMPARE_OP 8 (is)
20 RETURN_VALUE
>> 21 ROT_TWO
22 POP_TOP
23 RETURN_VALUE
Solution 3 - Python
From the documentation:
> 5.9. Comparisons > > Unlike C, all comparison operations in Python have the same priority, which is lower than that of any arithmetic, shifting or bitwise operation. Also unlike C, expressions like a < b < c have the interpretation that is conventional in mathematics:
comparison ::= or_expr ( comp_operator or_expr )*
comp_operator ::= "<" | ">" | "==" | ">=" | "<=" | "<>" | "!="
| "is" ["not"] | ["not"] "in"
Solution 4 - Python
True == False is False is a chained comparison, which means the same as (True == False) and (False is False). Since the first comparison (True==False) is false, the result of the chained comparison is False.