February 2011 - Haskell-Cafe

Type problem
by michael rice 26 Feb '11

26 Feb '11

Why? Shouldn't this work for any type a? Michael ========== f :: [a] -> [a] f l = do x <- l return x ========== *Main> :r [1 of 1] Compiling Main ( test.hs, interpreted ) Ok, modules loaded: Main. *Main> f "abcde" "abcde" *Main> f [1,2,3,4,5] [1,2,3,4,5] *Main> "abcde" >>= f <interactive>:1:12: Couldn't match expected type `Char' against inferred type `m b' In the second argument of `(>>=)', namely `f' In the expression: "abcde" >>= f In the definition of `it': it = "abcde" >>= f *Main>

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Simulating open datatypes with type families?
by Patrick Thomson 26 Feb '11

26 Feb '11

I'm currently working on a project that involves a C parser (using Parsec3) that needs to be dynamically extensible - i.e. end-users should be able to add new statement types, expression types, operators, and so on and so forth. Since Haskell ADT's are closed, I thought I would be able to simulate this by using type families, e.g.: {-# LANGUAGE TypeFamilies, DeriveDataTypeable #-} -- the statement type type CStatement = BreakStmt | CaseStmt CExpression CStatement -- rest of the statement types omitted here for brevity -- CStatement derives (Show, Eq, Typeable, Data) statement :: Parser CStatement -- how exactly this is parsed is not relevant, so its definition is also omitted data family StatementLike a data instance StatementLike CStatement = MkStmt CStatement -- okay, this is all well and good. let's create a new ADT, representing the Ruby/Perl-esque 'until' statement and add a StatementLike instance for it. data Until = UntilStmt CExpression CStatement deriving (Eq, Show, Typeable, Data) data instance StatementLike Until = MkUntil Until untilStatement :: Parser Until -- definition omitted -- Here's where it breaks down. I want to be able to have a function of type Parser (StatementLike a) so that I can pattern-match over the a in future code, but this function doesn't typecheck: GHC says that the function can't be that general - it expects either a Parser (StatementLike CStatement) or a Parser (StatementLike Until). newStatement :: Parser (CStatement a) newStatement = (MkUntil <$> untilStatement) <|> (MkStmt <$> statement) Am I expecting the wrong thing of type families? Are typeclasses the better way to do this? Is there any other way to simulate an open, extensible ADT, or am I totally barking up the wrong tree? Any assistance would be much appreciated. Please let me know if I need to provide more information. Thanks in advance, -- Patrick Thomson

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Auto elimination of MVars using a monad or monad transformer.
by Chris Dew 26 Feb '11

26 Feb '11

Hello, just like everyone else, I have a question about monads. I've read the tutorials, written one monad myself (not in this email), but I still consider myself a Haskell beginner. * Does GHC eliminate unneeded MVars during compilation? I'm expecting that it doesn't, as that would mean optimising away ForkIOs, which would be quite a thing to do. I've included example code below. * Is there a monad which allows their automatic elimination of MVars (or their creation only when necessary)? This would be similar to how the IO monad allows you to do purely functional things with a do block, using let. I've had a go at a lifting function, which wraps a pure function into an IO action which forever reads from one MVar and writes to another. What I'm looking for is some form of Monadic context in which many pure functions, MVar fillers and MVar consumers could be linked together, where only the necessary MVars remain (or were created) at compilation time. * Would this be a monad, or a monad transformer? * Can you specialise a monad transformer on a single base (in this case IO) so that you can use forkIO in the bind or return? Thanks, Chris. module Main ( main ) where import Control.Concurrent (forkIO, MVar, newEmptyMVar, putMVar, takeMVar, ThreadId, threadDelay) import Control.Monad (forever) stepA :: MVar String -> IO () stepA boxa = forever $ do line <- getLine putMVar boxa line stepB :: MVar String -> IO () stepB boxb = forever $ do line <- takeMVar boxb putStrLn line -- This simply wraps a string in brackets. bracket :: String -> String bracket x = "(" ++ x ++ ")" -- This lifts a function into an action which forever performs the function -- between the two MVars given. lft :: (a -> b) -> MVar a -> MVar b -> IO () lft f c d = forever $ do x <- takeMVar c putMVar d (f x) -- Just like C's main. main :: IO () main = do box <- newEmptyMVar box2 <- newEmptyMVar forkIO $ stepA box forkIO $ lft bracket box box2 forkIO $ stepB box2 threadDelay 10000000 -- Sleep for at least 10 seconds before exiting.

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Haskell platform installation failure on OS X 10.6.6 (intel)
by Eugene Kirpichov 26 Feb '11

26 Feb '11

Hello, Below is an experience report from Artem (cc'd) on a failed installation of Haskell platform on a Mac. OS: OS X 10.6.6 CPU: Intel Core i5 Platform downloaded from: http://lambda.haskell.org/hp-tmp/2010.2.0.0/haskell-platform-2010.2.0.0.i38… Sequence of actions: Downloaded the file, mounted the image, ran "Install new GHC", pressed OK until the following screen appeared http://i.imgur.com/GGdbg.png - there the Install button was disabled; pressing "change install location" did not help since only 1 option was available there, and it was already selected. After that, Artem tried to uninstall GHC, but unsuccessfully, since OS told that it was not installed. Then he tried to install the platform, and eventually got the following error around "moving files into place": http://i.imgur.com/YHb22.png I've also heard somewhat similar problem reports on installing haskell-platform on OS X from other people (I myself don't have it) - like being unable to install it without first installing XCode, etc. Can anyone help resolve this issue? I'm sure this is not expected behavior. What further information would help find the cause? -- Eugene Kirpichov Senior Software Engineer, Mirantis Inc. http://www.mirantis.com/

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Haskell Platform install instructions
by Michael Snoyman 26 Feb '11

26 Feb '11

Hi everyone, Some of the dependencies for Yesod (such as haskell-src) require happy. For the "Yesod in Five Minutes" page[1], I've given instructions to cabal install happy and then cabal install yesod. However, as Xavier Shay pointed out to me, this won't actually work on a vanilla install, since happy gets installed in the cabal bin folder, which is not by default on the PATH. Two questions: * Should we document this somewhere in the Haskell Platform install process? I'm sure many old-time users of cabal are well aware that they need ~/.cabal/bin in the PATH, but new users will not be. I would also like to update Yesod's documentation; does anyone have standard phrasing that is applicable to all operating systems? * Should we consider modifying the build tools to automatically look for build dependencies in the cabal bin folder? Michael [1] http://docs.yesodweb.com/five-minutes

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FFI: C-side object not destructed
by Yves Parès 26 Feb '11

26 Feb '11

Hello, I'm trying to use a C++ class in Haskell through C exports. It works all very well, except that when I call the function that deletes the object, it is still valid, I can still call methods on it. Here is my Haskell code: {-# LANGUAGE ForeignFunctionInterface #-} import Foreign import Foreign.C.Types newtype PKlass = PKlass (Ptr PKlass) foreign import ccall unsafe "Klass_Create" kCreate :: CInt -> CInt -> IO PKlass foreign import ccall unsafe "Klass_Destroy" kDestroy :: PKlass -> IO () foreign import ccall unsafe "Klass_GetX" kGetX :: PKlass -> IO CInt foreign import ccall unsafe "Klass_GetY" kGetY :: PKlass -> IO CInt foreign import ccall unsafe "Klass_AddKlass" kAdd :: PKlass -> PKlass -> IO PKlass main = do k <- kCreate 4 5 kDestroy k *kGetY k >>= print -- This shouldn't work* k' <- kCreate 2 8 k'' <- k `kAdd` k' kDestroy k'' *kGetY k'' >>= print -- This neither* So it is very basic, and I can't understand why the supposedly destroyed objects are still here. Enclosed is all the source code (C++ class, C exportation and the Haskell main hereabove). I compile it this way: ghc --make main.hs *.cpp -lstdc++

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Either instance of Monad?
by Yves Parès 26 Feb '11

26 Feb '11

Hello, When importing Control.Monad.Error, it shows me that (Either e) is an instance of MonadFix, but not an instance of Monad. Isn't there a problem?

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Fun with the ST monad
by Andrew Coppin 26 Feb '11

26 Feb '11

OK, so I had a function that looks like transform :: [Word8] -> [Word16] It works nicely, but I'd like to use mutable state inside. No problem! Use the ST monad. Something like transform :: [Word8] -> [Word16] transform xs = runST (work xs) where work :: [Word8] -> ST s [Word16] Ah, yes, well there is one *small* problem... If you do that, the function becomes too strict. The input list is being read from disk by lazy I/O. With the original implementation, the input file gets read at the same time as the output file is written. But runST returns nothing until the *entire* input has been compressed. So writing to disk doesn't start until the entire file has been slurped up into memory. Anybody have any hints on how to get around this? My first thought was to break the ST computation into several seperate pieces. But, /by design/, you cannot do this. And here's why: main = do let r = runST $ newSTRef 0 let x = runST $ n <- readSTRef r; writeSTRef r (n+1); return n let y = runST $ n <- readSTRef r; writeSTRef r (n*2); return n print x print y Now what the hell does this print out? Because it appears, logically, that it ought to vary depending on the order in which x and y are accessed - a clear violation of referential integrity. Fortunately of course, by a clever dodge, this code doesn't actually type-check. That's great, because it has undefined semantics. But it's also a problem, since it's exactly the sort of thing I'd like to do. Come to think of it, how the heck does lazy I/O manage this trick? How come accessing the elements of the list in the wrong order doesn't cause the wrong data to end up in each cell, if each cell is just using unsafePerformIO to read the next byte from a normal file handle?

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best way to deal with #defines when developing ffi's
by briand＠aracnet.com 25 Feb '11

25 Feb '11

Hi all, I'm working on an FFI and as usual there are lots of defines in the header files. What's the best way to make these available to the haskell code ? Ideally it could be done automagically, or at least pseudo-magically so that keeping up with changes to the .h wouldn't be too painful. I'm talking about simple constant defines, nothing tricky like structures or similar. Thanks, Brian

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Proxy for cabal
by Hauschild, Klaus (EXT) 25 Feb '11

25 Feb '11

Hi, Is it possible to configure a proxy server for cabal? When yes, how I configure it? Thanks.

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