Crc-16

Hello

I'm using Darrel's CRC routine to calculate a CRC but the results are not as expected.

Here is Darrel's routine -

Code:

---

CRC      VAR WORD
CRC_IN    VAR BYTE
Idx      VAR BYTE
CRC_Len  CON 16
CRC_Poly  CON $1021  ; x16 + x12 + x5 + 1

;-----CRC-16 -----------------------------------------------------
CRC16:
  CRC.HighByte = CRC.HighByte ^ CRC_IN
  FOR Idx = 0 to 7
    IF CRC.0(CRC_Len-1) THEN
      CRC = (CRC << 1) ^ CRC_Poly
    ELSE
      CRC = CRC << 1
    ENDIF
  NEXT Idx
RETURN

---

The eight bytes I'm using are "F2 AA 4C D4 60 F6 00 80"

The result I get is B2D5 rather than E23E as it should be.

Using an online calculator I get the same result, but if I tick the box "bit reflected", I get the correct result.

I'm sure I'm missing something really simple but I have tried many things without success.

Any Ideas?

There are many variations of CRC16.
Along with the different polynomial ... Some need a starting value in the CRC variable, some need a final XOR of the CRC result with another value, some reverse the bit order of the data, and some use combinations of those methods.

The variation you are looking for is called CRC16_CCITT_TRUE or CRC16_KERMIT, which reverses the bit order.

For this type, the CRC16: code is the same.
But when you are feeding it data, reverse the order of each byte.
Then reverse the order of the final result.

Code:

---

    ARRAYWRITE MyArray,[$F2,$AA,$4C,$D4,$60,$F6,$00,$80] ; Your Data


    CRC = 0
    FOR X = 0 TO 7
        CRC_IN = MyArray(X) REV 8
        GOSUB CRC16
    NEXT X
    CRC = CRC REV 16

---

The above gives the result ... $E23E

Hello Darryl

Thank you for that. It now looks fine.

I did try reversing the bytes but didn't think about reversing the result, so I guess I was about halfway to solving it.

You have saved me a lot of time - thanks again.

BR

Reg Smith
:D