[Haskell-beginners] Opaque (to me) compilation error

[Theodore Lief Gannon]

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 at 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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