Python 为什么 `True == False is False` 评估为 False？
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Why does `True == False is False` evaluate to False?
提问by raylu
I get some rather unexpected behavior on an expression that works with ==but not with is:
我在一个适用于==但不适用于 的表达式上得到了一些相当意外的行为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
采纳答案by jorgeca
Because in fact that's a chained comparison, so
因为实际上这是一个链式比较，所以
True == False is False
is equivalent to
相当于
(True == False) and (False is False)
This can be surprising in this case, but lets you write 1 <= x < 4unlike in other languages like C.
在这种情况下，这可能令人惊讶，但可以让您编写1 <= x < 4与 C 等其他语言不同的代码。
回答by BrenBarn
True == False is Falseis 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.
True == False is False是一个链式比较，这意味着与 相同(True == False) and (False is False)。由于第一次比较 ( True==False) 为假，因此链式比较的结果为假。
回答by SylvainD
From the documentation:
从文档：
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:
5.9. 比较
与 C 不同，Python 中的所有比较操作都具有相同的优先级，低于任何算术、移位或按位运算的优先级。同样与 C 不同的是，像 a < b < c 这样的表达式具有数学中的常规解释：
comparison ::= or_expr ( comp_operator or_expr )*
comp_operator ::= "<" | ">" | "==" | ">=" | "<=" | "<>" | "!="
| "is" ["not"] | ["not"] "in"
回答by Ashwini Chaudhary
From the docs:
从文档：
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).
x < y <= z 等价于 x < y 和 y <= z，除了 y 只计算一次（但在这两种情况下，当发现 x < y 为假时，根本不计算 z）。
In your case True == False is Falseis equivalent to True == False and False is Falseas the first condition is Falseso it short-circuits and return False.
在您的情况下True == False is False，相当于True == False and False is False第一个条件，False因此它短路并返回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
