Python - If/Elif statement (loops)

Question

Can someone tell me why I'm getting 4 as the outcome? Should 3 be printing instead because the code is satisfied in the If section.

a = 15

if a&5  == 0:
    print('3')
elif a% 5 == 0:
    print('4')

else:
    print('5')

Because 15&5 will evaluate to 5 and 5 never equals to 0. — adnanmuttaleb
– adnanmuttaleb, Commented Feb 18, 2020 at 7:09

timgeb · Accepted Answer · 2020-02-18 07:11:34Z

4

& is bitwise AND and works like this:

15 & 5

  1111
& 0101
------
  0101 != 0

answered Feb 18, 2020 at 7:11

timgeb

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Comments

Roger Lindsjö · Accepted Answer · 2020-02-18 07:11:27Z

3

You are comparing the bits of 15 with the bits of 5, resulting in the ones they have in common.

1111 & 101 => 101

so the result is 5, not 0.

answered Feb 18, 2020 at 7:11

Roger Lindsjö

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Comments

Mehdi · Accepted Answer · 2020-02-18 07:20:10Z

3

The "&" operator is a Bitwise AND so in your code it is like:

1111 & 0101 = 0101

That means

15 and 5 = 5

So the first condition can not be true and you'll get 4 in output because the reminder of 15/5 is 0.

edited Feb 18, 2020 at 7:20

answered Feb 18, 2020 at 7:14

Mehdi

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Comments

Sameer Roy · Accepted Answer · 2020-02-18 07:25:27Z

1

If you want to check LSB then you should do "and" with 1 and check the result if it's 1 then LSB is 0 else LSB is 1

answered Feb 18, 2020 at 7:25

Sameer Roy

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