Does anyone have a snippet for a "cyclic redundancy check"?

idbruce

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Dave Hein

PUB ComputeCRC(ptr, num)
  repeat num
    result ^= (byte[ptr++] << 8)
    repeat 8
      result <<= 1
      if result & $10000
        result ^= $1021
  result &= $ffff

idbruce

Thanks Dave

That was quick. Have you used this successfully?

I assume the ptr is to the data. And what is an example of a good "num" parameter?

Bruce

idbruce

Dave

If at all possible, could you please provide an example of use on both ends? Never mind, I think I got it.

Bruce

Mike Green

Bruce,
If you look at the code, you'll see that the CRC calculation loop is repeated "num" times, so "num" must be the byte count.

idbruce

Mike

Thanks for THAT information, I thought it was an arbitrary number, that would produce different hash results. WOW that would have had me going in a mental loop for a while.

Bruce

idbruce

LOL Did I say that right? It is a hash alogorithim is it not?

Dave Hein

Bruce,

The code that I posted will generate the stadard ITU 16-bit CRC for a block of "num" bytes starting at the address in "ptr". This routine assumes that the data is transmitted LSB first. If the data is transmitted MSB first a different routine would be needed, which would use a polynomial word of $8408 instead of $1021, which is the bit-reversed version of $8408. If you want to use this to generate a 16-bit hash code of a string you would do something like hash := ComputeCRC(@strbuf, strsize(@strbuf))

Dave

idbruce

Dave

Thanks for that information. I appreciate you sharing your snippet

Bruce