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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; }assert IsEven(-2);Now see, you need the other method. IsNegativeEven()
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(); }Nice, now we just need another function to find odd numbers, cause not all numbers are even you know.
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; }What'd be the result for
IsEven(1)?Stack overflow
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