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submitted 3 weeks ago by cm0002@lemmy.world to c/programmer_humor@programming.dev

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[-] sus@programming.dev 86 points 3 weeks ago* (last edited 3 weeks ago)

After working at blizzard for 51 years, I finally found an elegant solution by using the power of recursion

private bool IsEven(int number){
  if (number > 1) return IsEven(number - 2);
  if (number == 0) return true;
  if (number == 1) return false;
}

[-] elvith@feddit.org 39 points 3 weeks ago

assert IsEven(-2);

[-] lambda@programming.dev 14 points 3 weeks ago

Now see, you need the other method. IsNegativeEven()

[-] sus@programming.dev 9 points 2 weeks ago

We can avoid expensive branches (gasp) by using some bitwise arithmetic to achieve the so-called "absolute value", an advanced hacker technique I learnt at Blizzard. Also unlike c, c# is not enlightened enough to understand that my code is perfect so it complains about "not all code paths returning a value".

private bool IsEven(int number)
{
    number *= 1 - 2*(int)(((uint)number & 2147483648) >> 31);
    if (number > 1) return IsEven(number - 2);
    if (number == 0) return true;
    if (number == 1) return false;
    throw new Exception();
}

[-] UnknownSoul@programming.dev 2 points 2 weeks ago

Nice, now we just need another function to find odd numbers, cause not all numbers are even you know.

[-] Ebber@lemmings.world 12 points 3 weeks ago

I removed the tail recursion for you:

private book IsEven(int number) {
    if(number > 1) return IsEven(number - 2) == true;
    if(number == 0) return true; 
    if(number == 2) return false;
}