利用异或来判断输入数值的奇偶性,该如何解决

利用异或来判断输入数值的奇偶性

C# code

 /// <summary>    /// 利用异或来判断奇偶性    /// 先把十进制转换为二进制后再利用二进制位逐个异或来判断结果奇偶性    /// h：当h为０时，flag为true则为奇数，反之亦然    /// </summary>    class Program    {        static void Main(string[] args)        {             int a,h;             bool flag=false;            string b;            char[] c = { };            while (true)            {                h = 0;                Console.WriteLine("请输入一个十进制数：");                try                {                    a = Convert.ToInt32(Console.ReadLine());                }                catch                {                    Console.WriteLine("输入不正确，正常退出！");                    break;                }                b = Convert.ToString(a, 2);                c = b.ToCharArray();                    for (int k = 0; k < c.Length - 1; k++)                {                    h ^= c[k] ^ c[k + 1];                    if (h == 0)                    {                        flag = false;                    }                    else                    {                        flag = true;                    }                }                if (flag)                {                    Console.WriteLine("偶数\n");                }                else                {                    Console.WriteLine("奇数\n");                }            }            Console.ReadKey();        }    }

C# code

b = Convert.ToString(a, 2);c = b.ToCharArray();for (int k = 0; k < c.Length - 1; k++){    h ^= c[k] ^ c[k + 1];}if (h != 0){    Console.WriteLine("偶数\n");}else{    Console.WriteLine("奇数\n");}
[解决办法]
while (true)
{
      string s = Console.ReadLine();
      int x = 0;
      if (!int.TryParse(s, out x))
          continue;
      Console.WriteLine((x ^ (x + 1)) == 1 ? "偶数" : "奇数");
}