You're missing IO in the type declaration, which I believe means that do block is running in the Id monad -- by inference, Id ByteString.

On Fri, Nov 13, 2015 at 8:41 PM, Dan Stromberg <strombrg@gmail.com> wrote:

In the following code:

prefix_md5 :: String -> Data.ByteString.ByteString
prefix_md5 filename = do
    let prefix_length = 1024
    file <- System.IO.openBinaryFile filename System.IO.ReadMode :: (IO System.IO.Handle)
    data_read <- Data.ByteString.hGet file prefix_length :: (IO Data.ByteString.ByteString)
    _ <- System.IO.hClose file
    let hasher = Crypto.Hash.MD5.init :: Crypto.Hash.MD5.Ctx
    let hasher2 = Crypto.Hash.MD5.update hasher data_read :: Crypto.Hash.MD5.Ctx
    let digest = Crypto.Hash.MD5.finalize hasher2 :: Data.ByteString.ByteString
    return digest :: (IO Data.ByteString.ByteString)

I get the error:

Md5s.hs:13:5:
    Couldn't match type `IO Data.ByteString.ByteString'
                  with `Data.ByteString.ByteString'
    Expected type: IO System.IO.Handle
                   -> (System.IO.Handle -> IO Data.ByteString.ByteString)
                   -> Data.ByteString.ByteString
      Actual type: IO System.IO.Handle
                   -> (System.IO.Handle -> IO Data.ByteString.ByteString)
                   -> IO Data.ByteString.ByteString
    In a stmt of a 'do' block:
      file <- System.IO.openBinaryFile filename System.IO.ReadMode ::
                IO System.IO.Handle

How should I interpret that error to solve this kind of problem on my own in the future?  I don't see where the line in question does anything with ByteString's!

How might I correct this function to eliminate the error?

Thanks!

--

Dan Stromberg

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