RFID协议中CRC的计算

我唯一的办法是实现逐位算法(TMS37157数据表图52)。

UINT16 rfid_get_crc(const UINT8 * data, INT8 size)
{
 static BOOL lsb;
 static BOOL rxdt;
 static UINT16 crc;
 static UINT8 bits;
 static UINT8 byte;
 static UINT8 i;
 const UINT16 RFID_CRC_INIT = 0x3791;

 crc = RFID_CRC_INIT;
 for (i=0; i<size; i++)
 {
  bits = 8;
  byte = data[i]; // Next byte
  while (bits --> 0)
  {
   lsb = crc & 1; // Store LSB
   crc >>= 1; // Shift right 1 bit
   rxdt = byte & 1;
   if (rxdt)
    crc |= 0x8000; // Shift in next bit
   if (lsb) // Check stored LSB
    crc ^= 0x8000; // Invert MSB
   if (0x8000 == (crc & 0x8000)) // Check MSB
    crc ^= 0x0408; // Invert bits 3 and 10
   byte >>= 1; // Next bit
  }
 }
 return crc;
}

更紧凑的实现(伪代码)是:

// Least significant bit first (little-endian)
  // x^16+x^12+x^5+1 = 1000 0100 0000 1000 (1) = 0x8408
  function crc(byte array string[1..len], int len) {

     //Other RFID tags I have seen use initialization of 0x0000:
     //rem  := 0x3791;

     rem  := 0x3791;

      for i from 1 to len {
         rem  := rem xor string[i]
         for j from 1 to 8 {   // Assuming 8 bits per byte
              if rem and 0x0001 {   // if rightmost (most significant) bit is set
                 rem  := (rem rightShift 1) xor 0x8408
             } else {
                 rem  := rem rightShift 1
             }
         }
     }
     // A popular variant complements rem here
      return rem

这可以在以下代码片段5中找到:

http://en.wikipedia.org/wiki/Computation_of_cyclic_redundancy_checks