• λλλ
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      fedilink
      1311 days ago

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

      • @[email protected]
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        fedilink
        711 days 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();
        }
        
        • @[email protected]
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          fedilink
          110 days ago

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