Hi
hPutInt h = hPutStr h . map chr . map (0xff .&.) . take 4 . iterate (`shiftR` 8)hGetInt h = replicateM 4 (hGetChar h) >>= return . foldr (\i d -> i `shiftL` 8 .|. ord d) 0 This of course assumes that a Char is read/written as a single low-order byte without any conversion. But you'd have to assume a lot more if you started messing with pointers. (Strange, somehow I get the feeling, the above is way too easy to be the answer you wanted.)
It's exactly the answer I was hoping for! Thanks Neil _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
