Calculating CRC for int32 data format using the Modbus protocol

Question

I am connecting to a device using the modbus protocol. I need to obtain 3 values from the machine. The first value is of the data format int16 and when I send an example byte array:

static byte[] hz = new byte[] { (byte) 0x01, (byte) 0x03, (byte) 0x00,
        (byte) 0x33, (byte) 0x00, (byte) 0x01 };

and use a CRC calculation method I obtained from a previous question I asked on the subject.

    // Compute the MODBUS RTU CRC
private static int ModRTU_CRC(byte[] buf, int len)
{
  int crc = 0xFFFF;

  for (int pos = 0; pos < len; pos++) {
    crc ^= (int)buf[pos];          // XOR byte into least sig. byte of crc

    for (int i = 8; i != 0; i--) {    // Loop over each bit
      if ((crc & 0x0001) != 0) {      // If the LSB is set
        crc >>= 1;                    // Shift right and XOR 0xA001
        crc ^= 0xA001;
      }
      else                            // Else LSB is not set
        crc >>= 1;                    // Just shift right
    }
  }

    // Note, this number has low and high bytes swapped, so use it accordingly (or swap bytes)
    return crc;  
    }

I can recieve a response. However, the other two values are of the int32 data format and do not return a reply when I use this method. To help troubleshoot I am using a program called Realterm. to fire off the commands as well. I use it to append a Modbus 16 CRC to the end of the byte stream and send it, this works for all three and returns the desired reply. Is this a case of the data format not working with this specific calculation formula? Whats the difference between CRC16 and Modbus16?

Answer 1

Modbus16 is a CRC16. CRC calculations have several parameters:

• the bit width, in this case 16
• the polynomial, in this case 0xA001
• the initial value,in this case 0xFFFF
• the bit order
• whether the final CRC is inverted with an XOR.

There are quite a number of CRC16s defined, with different values for these parameters, and this appears to be one of them. See the Wikipedia article on Cyclic Redundancy Checks for more informaton.

Answer 2

class Obliczenia {

short POLYNOM = (short) 0x0A001;
short[] TAB = {2,3,8,0x13,0x88,1,0x90,0,0x3c,2,0};
short crc = (short) 0xffff;
short CRC_LByte,CRC_HByte;
  public Obliczenia() {
    for (short dana : TAB) {
        crc= CRC16( crc, dana);
    }
    System.out.println("KOD CRC="+Integer.toHexString(crc&0xffff));
    CRC_LByte=(short)(crc & 0x00ff);
    CRC_HByte=(short)((crc & 0xff00)/256);
     System.out.println(" W ramce CRC_LByte="+Integer.toHexString(CRC_LByte)+ "    CRC_HByte   "+Integer.toHexString(CRC_HByte));

}
short CRC16(short crct, short data) {
    crct = (short) (((crct ^ data) | 0xff00) & (crct | 0x00ff));
         for (int i = 0; i < 8; i++) {
        boolean LSB = ((short) (crct & 1)) == 1;
         crct=(short) ((crct >>> 1)&0x7fff);
        if (LSB) {
            crct = (short) (crct ^ POLYNOM);
        }
    }
    return crct;
}

}