[Haskell-beginners] Word8, Word32, ByteString, Int,
daniel.is.fischer at web.de
Tue Oct 12 14:18:53 EDT 2010
On Tuesday 12 October 2010 18:38:31, David McBride wrote:
> I'm writing a pcap sniffer, and I have IP addresses from two different
> libraries that are equivalent, but typed differently.
> Data.IP.IPv4 Data.ByteString.Internal.ByteString
> Network.Info.IPv4 = Network.Info.IPv4 !GHC.Word.Word32
> Both are probably identical, but I cannot for the life of me figure out
> how to get from one to the other. Bytestring has the ability to take
> arrays of Word8, how do I split Word32 into Word8's?
If you have the binary package and know the endianness of the IP addresses,
the simplest way to convert between those is
conv1 :: Network.Info.IPv4 -> Data.IP.IPv4
conv1 (Network.Info.IPv4 w32) =
Data.IP.IPv4 (runPut $ putWord32be w32)
-- or putWord32le
conv2 :: Data.IP.IPv4 -> Network.Info.IPv4
conv2 (Data.IP.IPv4 bs) = Network.Ifo.IPv4 (runGet $ getWord32be)
-- or getWord32le
Alternatively, you can convert a Word32 to [Word8] per
octets :: Word32 -> [Word8]
octets w =
[ fromIntegral (w `shiftR` 24)
, fromIntegral (w `shiftR` 16)
, fromIntegral (w `shiftR` 8)
, fromIntegral w
for big-endian conversion (for little-endian, reverse the list ;)
And a list of Word8 to a Word32 per
fromOctets :: [Word8] -> Word32
fromOctets = foldl' accum 0
accum a o = (a `shiftL` 8) .|. fromIntegral o
(reverse or use foldr for little-endian conversion).
> I have also had this problem in the past with Word8's, Octets and Chars.
> Beyond this specific problem, what is a good strategy for figuring out
> how to do these conversions in the future? Sometimes I find a function
> somewhere, but I have never found a general way to deal with it.
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