两个字节数组byte[]怎么用C#实现异或运算？请帖代码，呵呵

Question

byte[]数组1：0x06 0x12 0x34 0x56 0xFF 0xFF 0xFF 0xFF  
byte[]数组2: 0x00 0x00 0x67 0x89 0x01 0x23 0x45 0x67  数组1和数组2进行异或运算，如下：       
               0x06 0x12 0x34 0x56 0xFF 0xFF 0xFF 0xFF           异或：0x00 0x00 0x67 0x89 0x01 0x23 0x45 0x67         结果为：0x06 0x12 0x53 0xDF 0xFE 0xDC 0xBA 0x98请各位大大将代码贴出来，帮助下小弟。呵呵

Answer 1

是对两个等长的byte数组，逐元素进行XOR运算，然后把结果保存进另一个数组吗？如果是，那就是一个循环啊byte[] result = new byte[array1.Length];
for(int i =0; i<array1.Length; i++)
    result[i] = array1[i] ^ array2[i];

Answer 2

var data1 = new[]{ 0x06, 0x12, 0x34, 0x56, 0xFF, 0xFF, 0xFF, 0xFF };
var data2 = new[]{ 0x00, 0x00, 0x67, 0x89, 0x01, 0x23, 0x45, 0x67 };
var data3 = new byte[data1.Length];
for(int i=0; i<data1.Length; i++)
    data3[i] = data1[i] ^ data2[i];

Answer 3

注意异或操作将byte隐式转换为int，需要再显示转换为byte            byte[] arr1 = { 0x06, 0x12, 0x34, 0x56, 0xFF, 0xFF, 0xFF, 0xFF };
            byte[] arr2 = { 0x00, 0x00, 0x67, 0x89, 0x01, 0x23, 0x45, 0x67 };
            byte[] arrXor = new byte[arr1.Length];            for (int i = 0; i < arr1.Length; i++)
            {
                arrXor[i] = (byte)(arr1[i] ^ arr2[i]);
                Console.WriteLine("0x{0:x}", arrXor[i]);
            }

Answer 4

byte[] bs1 = {0x06, 0x12, 0x34, 0x56, 0xFF, 0xFF, 0xFF, 0xFF };
            byte[] bs2 = {0x00, 0x00, 0x67, 0x89, 0x01, 0x23, 0x45, 0x67};
            byte[] bs3 = new byte[bs1.Length];
            for (int i = 0; i < bs1.Length; i++)
            {
                bs3[i] = (byte)(bs1[i] ^ bs2[i]);
            }

Answer 5

var 结果=数组1.Zip(数组2,(b1,b2)=>b1^b2).ToArray();