Why 'and' and 'or' behaves differently with 0, 1, True, False?

Question

Below is the scenario.

>>> False and 0
=> False
>>> 0 and False
=> 0

Both the conditions are same, but why is it returning different results?

>>> 0 or False
=> False
>>> False or 0
=> 0

Similarly, both the conditions should return same results, but why are they different?

The same applies in case of True and 1

@ΦXocę웃Пepeúpaツ Is there a specific order of execution when we use both bool and int? — radh, Jun 09 '17 at 11:29
@ΦXocę웃Пepeúpaツ booleans are `int`s in Python, try `isinstance(False,int)` — Chris_Rands, Jun 09 '17 at 11:29
As the dupe says " a and b returns b if a is True, else returns a. a or b returns a if a is True, else returns b. " — Chris_Rands, Jun 09 '17 at 11:31
@ΦXocę웃Пepeúpaツ `isinstance(False,bool)` and `isinstance(False,int)` returns same. Is it like, bool values are interpreted as integers? — radh, Jun 09 '17 at 11:34

blue_note · Accepted Answer · 2017-06-09T11:34:39.027

1

In python, theand, or operators do not return boolean. They return the last thing evaluated. Since they are short circuit operators, the last thing that need to be evaluated for the expression 0 and False, is0. Similarly, for 0 or False, the last thing that needs to be checked is the second operand, False.

From the python documentation:

x or y: if x is false, then y, else x
x and y: if x is false, then x, else y
not x: if x is false, then True, else False

edited Jun 09 '17 at 11:34

answered Jun 09 '17 at 11:27

blue_note

27,712
9
72
90

1

if that where true then doing ***False and 0*** should return 0 and not False! – ΦXocę 웃 Пepeúpa ツ Jun 09 '17 at 11:29
1

No, it should return False, since the `and` has already failed, it does not need to check 0. – blue_note Jun 09 '17 at 11:30
OP is showing us that ***False and 0*** returns in fact ***False*** – ΦXocę 웃 Пepeúpa ツ Jun 09 '17 at 11:31
@ΦXocę웃Пepeúpaツ: Again no, it returns False. The OP asked for `False OR 0` – blue_note Jun 09 '17 at 11:32
Nope, the OP's snippet for `False and 0` does yield `False` as expected - it's `False or 0` that yields `0` (as expected again). – bruno desthuilliers Jun 09 '17 at 11:33

score 0 · Answer 2 · answered Jun 09 '17 at 11:33

Python conditional checking uses Short-Circuit Evaluation. This means the following:

False and 0

It is executing an and statement, so both elements must evaluate to 1. Since False doesn't, it does not check for the second one and it returns False. This also applies for 0 and False, but since 0 is the first element it returns 0.

False or 0

In this case it is performing an or evaluation so one of the elements must evaluate to 1 (or True). Since False doesn't, it checks for the second operator (0) which neither evaluates to 1, so it returns it. The same applies for 0 or False.

Here you have another approach that clears out any doubt. Using 0 or "Text" it returns "Text" as it evaluates to True.

score 0 · Answer 3 · answered Jun 09 '17 at 11:39

or is a short-circuit operator, so it only evaluates the second argument if the first one is false. In the case, False or 0, first argument is evaluated to False, then second argument is evaluated, which is 0, and returns 0. This is valid for both the cases(0 or False too).

and is also a short-circuit operator, so it only evaluates the second argument if the first one is true. In the case False and 0, the first argument is evaluated to False, then the second is not evaluated and returns the first argument and vice versa.

For more clarification, refer the docs.

Why 'and' and 'or' behaves differently with 0, 1, True, False?

3 Answers3