196

Using Haskell you can write it way more concise:

iseven :: Int -> Bool
iseven 0 = True
iseven 1 = False
iseven 2 = True
iseven 3 = False
iseven 4 = True
iseven 5 = False
iseven 6 = True
iseven 7 = False
iseven 8 = True
...

However, we can be way smarter by only defining the 2 base cases and then a recursive definition for all other numbers:

iseven :: Int -> Bool
iseven 0 = True
iseven 1 = False
iseven n = iseven (n-2)

It's having a hard time with negative numbers, but honestly that's quite a mood

=if((number/2)-round(number/2,0)=0,true,false)

so did someone draw this by hand or was it a filter

a wise programmer knows to always ask the question "can i solve this problem in python using metaprogramming?" in this instance, the answer is yes:

def is_even(n: int):
    s = "def is_even_helper(number: int):\n"
    b = True
    for i in range(0, abs(n)+2):
        s += f"\tif (abs(number) == {i}): return {b}\n"
        b = not b
    exec(s)
    return locals().get("is_even_helper")(n)

oh of course there is

https://www.npmjs.com/package/is-even

(do take a look at the download stats)

I thought they were going to turn into Saddam Husseins.

Zero people in this post get the YanDev reference

if (!(number & 1))

just check the least significant bit smh my head

https://codegolf.stackexchange.com/q/275739/88192

return true

is correct around half of the time