• noddy@beehaw.org
    1·
    2 years ago

    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.");
    }
}
      • noddy@beehaw.org
        1·
        2 years ago

        I know an even better way. We can make it run in O(1) by using a lookup table. We only need to store 2^64 booleans in an array first.

  • neidu@feddit.nl
    1·
    2 years ago

    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” }

      • Chobbes@lemmy.world
        0·
        2 years ago

        You know, I was going to let this slide under the notion that we’re just ignoring the limited precision of floating point numbers… But then I thought about it and it’s probably not right even if you were computing with real numbers! The decimal representation of real numbers isn’t unique, so this could tell me that “2 = 1.9999…” is odd. Maybe your string coercion is guaranteed to return the finite decimal representation, but I think that would be undecidable.

        • backgroundcow@lemmy.world
          1·
          2 years ago

          Ackchyually-- IEEE 754 guarantees any integer with absolute value less than 2^24 to be exactly representable as a single precision float. So, the “divide by 2, check for decimals” should be safe as long as the origin of the number being checked is somewhat reasonable.

  • SpeakinTelnet@programming.dev
    1·
    2 years ago
    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)