Note that the analysis in my previous message is based on a trick I saw a year ago and clearly don't properly remember or understand, because there is no "Id" monad and the *Identity* monad is apparently not in base (though you might still have it there, it's in both mtl and transformers). But the bit about your type signature still stands. I'm quite certain of that part. :) On Fri, Nov 13, 2015 at 9:24 PM, Theodore Lief Gannon wrote:

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 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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On Sat, Nov 14, 2015 at 10:52 AM, Kim-Ee Yeoh wrote:

What we're seeing is special handling for stage 3: reporting errors in a do-block. Sometimes it helps, sometimes it hinders. For instance, the derivation of this signature is confusing indeed:

IO System.IO.Handle -> (System.IO.Handle -> IO Data.ByteString.ByteString) -> Data.ByteString.ByteString

To elaborate, the error message comes from the compiler choking on the monadic bind (>>=) that reveals itself in the desugared code. The (>>=) results from desugaring file <- openBinaryFile filename ReadMode into openBinaryFile filename ReadMode >>= \file -> ... The compiler knows the type signature of (>>=), which wants to see IO Handle -> (Handle -> IO ByteString) -> IO ByteString But the signature of the top-level function insists on IO Handle -> (Handle -> IO ByteString) -> ByteString This is what's the compiler's trying to say. -- Kim-Ee