You could do this in one line…

By removing all the linebreaks.

Programming humor on reddit used to be excellent bits like this but then it devolved into new learners jumping straight to the irony they didn’t understand and flooded the sub with nonsense.

I miss these bits.

btw it does get easier

import math
def is_even(num):
    if num in [i for i in range(1000) if float(i)/2.0 == math.floor(float(i)/2.0)]:
        print("true")
    else:
        print("false")

Obviously one would need to increase the range for bigger numbers but this code is optimized.

My solution in perl back in the day when I was a teenage hobbyist who didn’t know about the modulus operator: Divide by 2 and use regex to check for a decimal point.

if ($num / 2 =~ /\./) { return “odd” }
else { return “even” }

def is_even(n):
    match n:
        case 1:
            return False
        case 0:
            return True
        # fix No1
        case n < 0:
            return is_even(-1*n)
        case _:
            return is_even(n-2)

I know how to fix this!

bool IsEven(int number) {
    bool even = true;
    for (int i = 0; i < number; ++i) {
        if (even == true) {
            even = false;
        }
        else if (even == false) {
            even = true;
        }
        else {
            throw RuntimeException("Could not determine whether even is true or false.");
        }
    }

    if (even == true) {
        return even ? true : false;
    }
    else if (even == false) {
        return (!even) ? false : true;
    }
    else {
        throw RuntimeException("Could not determine whether even is true or false.");
    }
}

This could be optimized by using a recursive function.

I would love it if someone edited this example and posted it with two statements near the end that are reversed, implying inconsistent behaviour at random in the list ahead, seemingly making this solution less inefficient.

modulo

pseudocode:

if number % 2 == 0
  return "number is even" (is_num_even = 1 or true)
else
  return "number is odd" (is_num_even = 0 or false)

plus you’d want an input validation beforehand

Of course there’s an easier way. Just integrate the state of the art API dedicated for this exact problem. https://isevenapi.xyz/