Logical operator binding of AND and OR in Python

Question

Yesterday while constructing a conditional statement I ran into what seems to me to be a strange precedence rule. The statement I had was

if not condition_1 or not condition_2 and not condition_3:

I found that

if True or True and False:
    # Evaluates to true and enters conditional

In my mind (and from previous experience in other languages) the and condition should have precedence as the statement is evaluated - so the statement should be equivalent to

if (True or True) and (False):

But in actual fact it is

if (True) or (True and False):

Which seems odd to me?

what is the question?

Francesco Nazzaro
– Francesco Nazzaro

2016-09-09 08:59:57 +00:00
Commented Sep 9, 2016 at 8:59 — Francesco Nazzaro
– Francesco Nazzaro, Commented Sep 9, 2016 at 8:59
or has a lower precedence than and

Łukasz Rogalski
– Łukasz Rogalski

2016-09-09 09:00:38 +00:00
Commented Sep 9, 2016 at 9:00 — Łukasz Rogalski
– Łukasz Rogalski, Commented Sep 9, 2016 at 9:00

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

From python wiki :

The expression x and y first evaluates x; if x is false, its value is returned; otherwise, y is evaluated and the resulting value is returned.

The expression x or y first evaluates x; if x is true, its value is returned; otherwise, y is evaluated and the resulting value is returned.

So in your case

if True or True and False:

Since interpreter encounters True before or it doesn't continue to and expression and evaluates to True

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Sep 9, 2016 at 9:00

garg10may

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Comments

Csaba Benko · Accepted Answer · 2016-09-09 09:08:03Z

0

As you said and has a higher preference, so your expression with parenthesis will be exactly what you said:

if (True) or (True and False)

Since True or (anything) will be True regardless anything here, the result will be True. Even if anything would be evaluated, which is not in a shortcut evaluation.

answered Sep 9, 2016 at 9:08

Csaba Benko

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