True & False is giving True in Python. Why?

Asked Jul 09 '19 at 08:11

Active Jul 09 '19 at 08:25

Viewed 65 times

I was doing a HandON with below Code:

for i in range(10,16):
    if(i%5==0 & i%3==0):
        print(i, 'both')
    elif i%5==0:
        print(i,'divisible by 5')
    elif i%3==0:
        print(i,'divisible by 3')

When I am executing this Code, it is giving Output as :

10 both
12 divisible by 3
15 both

Why is it giving 10 as divisible by both? If it is Bit-wise operator, then 'if' will have '0001' and '0000' for value 10, which should result in '0000'. If I use 'and' instead of '&', it is giving correct answer.

edited Jul 09 '19 at 08:25

asked Jul 09 '19 at 08:11

Gopesh

5

`if((i%5==0) & (i%3==0))` – anky Jul 09 '19 at 08:13
5

`True & False` is not `True` in python. (But then, what's going on here you ask?! Clearly, you're making an assumption that doesn't hold true. in this case, it's operator precedence. try brackets around the conditions before using `&`) – Paritosh Singh Jul 09 '19 at 08:14
3

it evaluates the condition like this: `if i%5== (0 & (i%3==0)):` – Gábor Fekete Jul 09 '19 at 08:14
3

@trotta Not a duplicate. OP is aware that & is bit-wise – DeepSpace Jul 09 '19 at 08:15
The title of this question is under a wrong assumption; `python -c "print(True&False)"` apparently prints `False`. – Ignatius Jul 09 '19 at 08:17
@ParitoshSingh Thanks for the detailed explanation. – Gopesh Jul 09 '19 at 08:25

True & False is giving True in Python. Why?

0 Answers0