why does the code > if 3 == 1 or 2: print('A') else: print('B') < print A but not B

Question

I recently started learning my first programming language and i couldn't understand why this code gave me A as the answer and not B

if 3 == 1 or 2:
    print('A')
else:
    print('B')

`3 == 1` is False, but `2` evaluates to True, so `False or True` is `True`. — mch, Nov 19 '19 at 13:05
I think you tried to compare like this: `if 3 == 1 OR 3 == 2: print('A') else: print('B')` In that case, you'd get `A` — DasSaffe, Nov 19 '19 at 13:06
related to https://stackoverflow.com/questions/16679272/priority-of-the-logical-statements-not-and-or-in-python — aloisdg, Nov 19 '19 at 13:11

score 1 · Accepted Answer · answered Nov 19 '19 at 13:10

You check for the two conditions 3==2 and 2. 2 evaluates to true, so you get A as output.

If you want to check 3 for beeing 1 or 2, you have to do it like this:

if 3 == 1 or 3 == 2: print('A') else: print('B')

See e.g. Do not understand why if condition evaluates to True when variable is integer for details why 2 is true, especially this answer.

score 0 · Answer 2 · answered Nov 19 '19 at 13:09

0

In python it could be read as if 3==1 or bool(2) == True where the first condition if False and the second is True because bool(2) is True

answered Nov 19 '19 at 13:09

Dmitrii

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score 0 · Answer 3 · answered Nov 19 '19 at 13:10

You have a wrong assumption that 3 == 1 or 2 means "is 3 equal to 1 or 2?", but in reality it's rather "is either 3 equal to 1, or does 2 evaluate to True" and since 2 is non-zero value it always evaluates to True.

More specifically, 3 == 1 or 2 expression evaluates to False or 2 then False or True and finally True.

If you want the expression that matches your original expectation, you should write 3 in [1, 2] which checks whether 3 is in one of the allowed values in the list.

why does the code > if 3 == 1 or 2: print('A') else: print('B') < print A but not B

3 Answers3