Re: [Haskell-cafe] Endian conversion

well, fastest conversion to compute could be an assembler-command, but if we don't use that, it could be converted via Foreign.Storable and sth like the following: (i did not test it, and i hope, TH works like this...) data (Integral a) => BigEndian a = BigEndian a deriving (Eq,Ord,Enum,...) be = $( (1::CChar)/=(unsafePerformIO $ with (1::CInt) $ peekByteOff `flip` 0) ) :: Bool instance (Storable a) => Storable (BigEndian a) where sizeOf (BigEndian a) = sizeOf a alignment (BigEndian a) = alignment a peek = if be then peek0 else peekR where peek0 (BigEndian a) = peek a peekR = peekByteOff `flip` 0 peekByteOff = if be then peekByteOff0 else peekByteOffR where peekByteOff0 (BigEndian a) = peekByteOff a peekByteOffR (BigEndian a) i = peekByteOff a (sizeOf a - 1 - i) ...poke... - marc Tomasz Zielonka wrote:

On 10/3/05, Joel Reymont wrote:

...

Folks,

Are there any endian conversion routines for Haskell? I'm looking to build binary packets on top of NewBinary.Binary but my data is coming in little-endian whereas I'll need to send it out big endian.

...

From your question I assume you want functions like htonl / ntohl. I think the cleanest approach is to always have yours Ints, etc in host order, and place the endianness stuff in serialization / deserialization code, ie. on the Number <-> Byte sequence boundary.

Having htonl/ntohl as pure functions in Haskell would be a bit ugly, because they would be defined differently on different platforms, and putting them in the IO monad would make them barely usable.

Best regards Tomasz

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