Programmer Humor

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Infallible Code (infosec.pub)

submitted 2 weeks ago by cm0002@lemmy.world to c/programmer_humor@programming.dev

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[–] sus@programming.dev 87 points 2 weeks ago* (last edited 2 weeks ago) (9 children)

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 40 points 2 weeks ago (3 children)

assert IsEven(-2);

[–] lambda@programming.dev 14 points 2 weeks ago (1 children)

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

[–] sus@programming.dev 9 points 2 weeks ago (1 children)

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.

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