一个CRC校验的问题

Question

CRC校验至少是2个字节吧，也有4个字节的
看你的接收方协议而定

Answer 1

如果是SUM校验,分1个字节和2个字节
如果是CRC校验,只分为CRC16和CRC32,没有CRC8给你段代码
        public void GetCheck(int ByteLength, ref byte[] ombuffer, string CheckMode)
        {
            int i, j = 0;
            byte hi, lo, c1, c2;
            switch (CheckMode)
            {
                case "CRC16":
                    hi = 0xff;
                    lo = 0xff;
                    for (i = 0; i < ByteLength; i++)
                    {
                        lo = (byte)(lo ^ ombuffer[i]);
                        for (j = 0; j < 8; j++)
                        {
                            c1 = lo;
                            c2 = hi;
                            lo = (byte)(lo >> 1);
                            hi = (byte)(hi >> 1);
                            if ((c2 & 0x01) != 0)
                            {
                                lo = (byte)(lo | 0x80);
                            }
                            if ((c1 & 0x01) != 0)
                            {
                                hi = (byte)(hi ^ 0xa0);
                                lo = (byte)(lo ^ 0x01);
                            }
                        }
                    }
                    ombuffer[ByteLength] = lo;
                    ombuffer[ByteLength + 1] = hi;
                    break;
                case "SUM":
                    lo = 0x00;
                    for (i = 0; i < ByteLength; i++)
                    {
                        lo += ombuffer[i];
                    }
                    ombuffer[ByteLength] = lo;
                    break;
            }        }        public bool CheckCRC(int ByteLength, byte[] imbuffer)
        {
            int i, j, k = 0;
            byte hi, lo, c1, c2;
            hi = 0xff;
            lo = 0xff;
            if (ByteLength < 4)
            {
                return false;
            }
            for (i = 0; i < (ByteLength - 2); i++)
            {
                lo = (byte)(lo ^ imbuffer[k]);
                for (j = 0; j < 8; j++)
                {
                    c1 = lo;
                    c2 = hi;
                    lo = (byte)(lo >> 1);
                    hi = (byte)(hi >> 1);
                    if ((c2 & 0x01) != 0)
                    {
                        lo = (byte)(lo | 0x80);
                    }
                    if ((c1 & 0x01) != 0)
                    {
                        hi = (byte)(hi ^ 0xa0);
                        lo = (byte)(lo ^ 0x01);
                    }
                }
                k++;
            }
            c1 = imbuffer[k];
            k++;
            c2 = imbuffer[k];            if ((c1 == lo) && (c2 == hi))
            {
                return true;
            }
            else
            {
                return false;
            }
        }