- Haskell-Cafe - Haskell.org

RE: infered type does not typecheck
by Simon Peyton-Jones 08 Jul '04

08 Jul '04

Duncan Oleg's recent message to Haskell-Café tackles your real problem, but I thought I'd just say what is going on with your apparent bug below. Here's a much simpler example: class Foo a b where to :: a -> b fro :: b -> a -- foo :: forall b1 b a. (Foo a b, Foo a b1) => b -> b1 foo x = to (fro x) Without the type signature, the program typechecks "fine", and GHCi reports the type in the comment. But if you uncomment the type signature, typechecking fails. Why? It's because the type is really ambiguous. 'fro' converts from b to a, and 'to' converts back... but the 'a' is entirely lost in the process. It's very like the classic (show (read x)) example. So the type foo :: forall b1 b a. (Foo a b, Foo a b1) => b -> b1 is really unusable, because there is nothing to constrain 'a'. When you add the type signature, GHC can't establish the connection between the 'a' needed as the intermediate type in the (to (fro x)) composition and the 'a' in the type signature. You might argue that GHC should reject foo in the first place, and you'd be right, really. But it's hard in general to reject exactly and only the ambiguous cases (given functional dependencies and all that jazz), and I have not worked hard on that. But I grant that it's rather counter-intuitive that adding a type signature that GHC itself deduces goes wrong! You're right to count it a bug. Simon | This was posted to the Haskell list earlier, | http://www.haskell.org//pipermail/haskell/2004-June/014286.html | but just to record what appears to be a typechecker bug. | | I have a function with this type: | buildUArray' :: (Ix i, MArray (array s) a (ST s)) => | (i,i) -> (i -> a) -> ST s (array s i a) | | Then I ask ghci (6.2.1 and 6.3 recent cvs) the type of the following | function: | | buildUArray bounds f = do arr <- buildUArray' bounds f | unsafeFreeze arr | | ghci says: | buildUArray :: forall a i b s array. | (MArray (array s) a (ST s), Ix i, IArray b a) => | (i, i) -> (i -> a) -> ST s (b i a) | | However if I add this annotation back in the script, ghc complains: | | Could not deduce (MArray (array1 s) a (ST s)) | from the context (MArray (array s) a (ST s), Ix i, IArray b a) | arising from use of `unsafeFreeze' at BuildArray.hs:25 | Probable fix: | Add (MArray (array1 s) a (ST s)) | to the type signature(s) for `buildUArray' | Or add an instance declaration for (MArray (array1 s) a (ST s)) | In the result of a 'do' expression: unsafeFreeze arr | In the definition of `buildUArray': | buildUArray bounds f | = do | arr <- buildUArray' bounds f | unsafeFreeze arr

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Re: [Haskell-cafe] Inferred type is not general enough
by Ivan Tihonov 08 Jul '04

08 Jul '04

Stefan Holdermans wrote: Huh. I'm really get stuck. Can someone write me working implementation of my crap?

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Re: [Haskell-cafe] Inferred type is not general enough
by Ben Lippmeier 08 Jul '04

08 Jul '04

Ivan.. Uh.. by 'works' I meant 'compiles' :) Here is a fixed version.. As I understand it, the 2 parameter class (Location loc => Packet p loc) means "loc is a Location, and types p and loc are related by class Packet" with just that information, if you try (as you did) something like $ source TestPacket Then you've given the compiler the type for 'p' (TestPacket), but it won't go the distance and decide that type for 'loc' is TestLocation.. even though that's the only instance you've given it. You can tell the compiler that "The type of 'p' sets the type of 'loc'" with a "functional dependency".. which is the "| p -> loc" annotation at the end of the class definition in the new version below. There is a paper which covers exactly this problem, in detail. I suggest you read it (I did). "Type classes with functional dependencies", Mark P. Jones, 2000 You can get it from his homepage at, http://www.cse.ogi.edu/~mpj/ Ben. Ivan Tihonov wrote: > It compiles well, but doesn't work for me :( > BTW: (email answers are not always real-time :) ) new version: -------------------------------------- class Location a where point :: a -> String class Location loc => Packet p loc | p -> loc where source :: p -> loc destination :: p -> loc size :: Num b => p -> b ------------------------------------------------------ data TestLocation = TestSource | TestDestination deriving (Show) data TestPacket = TestPacket ------------------------------------------------------ instance Location TestLocation where point a = "location" instance Packet TestPacket TestLocation where source p = TestSource destination p = TestDestination size p = 99

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Re: [Haskell-cafe] Folding function composition: L or R?
by Graham Klyne 07 Jul '04

07 Jul '04

At 14:24 07/07/04 +0200, Henning Thielemann wrote: >On Wed, 7 Jul 2004, Graham Klyne wrote: > > > [[ > > sss :: [ShowS] > > sss = map (++) ["abc","def","ghi","jkl"] > > > > ssr = foldr (.) id sss > > ssl = foldl (.) id sss > > > > sr = ssr "" > > sl = ssl "" > > > > -- sl == sr == "abcdefghijkl" > > -- is there any efficiency difference? > > ]] > > > > Conventional wisdom suggests that foldr is better, but I'm not sure if it > > really matters in this case. > >foldl performs worse, e.g. for infinite lists it will need infinite >time :-) Of course! And your answer makes me notice an important touchstone when dealing with such questions: "what happens with an infinite list?". Thanks! #g ------------ Graham Klyne For email: http://www.ninebynine.org/#Contact

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Folding function composition: L or R?
by Graham Klyne 07 Jul '04

07 Jul '04

I have a sequence of functions that I wish to fold together using function composition (.). Does it matter if I use foldl or foldr? e.g. [[ sss :: [ShowS] sss = map (++) ["abc","def","ghi","jkl"] ssr = foldr (.) id sss ssl = foldl (.) id sss sr = ssr "" sl = ssl "" -- sl == sr == "abcdefghijkl" -- is there any efficiency difference? ]] Conventional wisdom suggests that foldr is better, but I'm not sure if it really matters in this case. #g ------------ Graham Klyne For email: http://www.ninebynine.org/#Contact

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RE: [Haskell-cafe] readMVar and the devils
by Simon Marlow 07 Jul '04

07 Jul '04

On 06 July 2004 11:29, Conor T McBride wrote: > OK, here's what I want, what I do with it, and my attempt to deliver > it. But I'm not an expert, so please let me know if there's some > disastrous flaw... > > The signature/spec: > > type Hole x > > hole :: IO (Hole x) > -- returns a fresh empty hole for an x > askHole :: Hole x -> IO (Maybe x) > -- inspects the current contents (or none) of the hole; mustn't > -- block > tellHole :: Hole x -> x -> IO (Maybe x) > -- tries to write the hole, returns what was previously there; > -- if the hole was empty, the value supplied is installed > -- if the hole was full, it's unchanged (too late, pal!); > -- mustn't block > readHole :: Hole x -> IO x > -- blocks until the hole has been filled, then returns its > -- value > instance Eq (Hole x) > -- a kind of higher-level `pointer equality' The following should do the trick (but I wouldn't rule out any lurking race conditions, and perhaps there's an even simpler way to achieve this): data Hole x = Hole (MVar ()) -- wait on this for readHole (MVar (Maybe x)) -- always full, Just x <=> hole contains x instance Eq (Hole x) where (Hole m1 _) == (Hole m2 _) = m1 == m2 hole = do wait <- newEmptyMVar val <- newMVar Nothing return (Hole wait val) askHole (Hole wait val) = readMVar val tellHole (Hole wait val) x = do modifyMVar val $ \mb -> case mb of Nothing -> do putMVar wait (); return (Just x, Nothing) Just y -> return (Just y, Just y) readHole (Hole wait val) = do readMVar wait mb <- readMVar val return (fromJust mb) Cheers, Simon

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problem with hsql on windows
by Shanon Fernald 07 Jul '04

07 Jul '04

I'm using: GHC 6.2.1 HSQL 1.4 on a Windows XP machine. I'm trying to the compile the HSQL example using the eclipse Haskell plug-in, but I get the following error message: Chasing modules from: Main.hs Compiling Queries ( ./Queries.hs, C:\workspace\dbtest\out/Queries.o ) ./Queries.hs:3: Failed to load interface for `Database.HSQL': Bad interface file: C:/ghc/ghc-6.2.1/imports/Database/HSQL/GHC/imports/Database/HSQL.hi mismatched interface file versions: expected 6021, found 6020 What it appears is that my HSQL has an interface matching GHC 6.2 maybe? I just downloaded the HSQL 1.4 from the sourceforge site. The windows exe version is HSQL-1.4.exe. Is this the correct version of HSQL? Am I matching the GHC and HSQL versions correctly? If so, do you know what else I might be doing wrong? I am able to compile other programs with no problem. Thanks for help. It is fun to learn a new language and my Haskell books are on order :)

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Re: Roman Numerals and Haskell Syntax abuse
by George Russell 07 Jul '04

07 Jul '04

Christian Maeder wrote (snipped) > btw, now 127: > roman=(!5);n!a|n<1=""|n>=t=s!!(a+1):(n-t)!a|c>=t=s!!(2*d):c!a|1>0=n!(a-1)where d=a`div`2;s="ivxlcdm";c=10^d+n;t=5^(d+1)*2^(a-d) This is amazing! When I posted my first attempt of 181 characters I thought it might be minimal.

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Random Numbers for the beginner ?
by Crypt Master 06 Jul '04

06 Jul '04

Hi I have tried I swear, even googled for 45 minutes, but I cant seem to get random numbers working. In the documentation is has: rollDice :: IO Int rollDice = getStdRandom (randomR (1,6)) But if I type "getStdRandom (randomR (1,6))" into hugs in the context of module which imports Random, I get get errors. ERROR - Unresolved overloading *** Type : (Random a, Num a) => IO a *** Expression : getStdRandom (randomR (1,6)) Roll dice takes no parameters and returns an IO Int. So in thoery (mine at least ;-) ) running this as an expresion should work. I should get an IO Int back from the interpreter ? So I added RollDice to my module. This doesnt error, but it doesnt return anything except blank spaces: HasGal> rollDice HasGal> Integers or nums should automatically have show correct? So this should show me something ? Ultimatly I want to get randomRs infinite list working so I can build randNums = (take (length popList) [1..]) where the length of pop list is how many random numbers I want. My code works as it, just need to replace [1..] with some random numbers. Thanks, C _________________________________________________________________ Add photos to your e-mail with MSN 8. Get 2 months FREE*. http://join.msn.com/?page=features/featuredemail

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Re: [Haskell-cafe] Random Numbers for the beginner ?
by Jon Cast 06 Jul '04

06 Jul '04

sholderm(a)students.cs.uu.nl wrote: <snip> > Use > > (getStdRandom (randomR (1,6))) :: IO Int This much is fine, but... > instead. > > \begin{code} > module Main where > import Random > > > main :: IO () > main = do n <- (getStdRandom (randomR (1,6))) :: IO Int > ; print n > \end{code} Layout error: the semicolon must be indented further than the `n' is. Furthermore, the semicolon is unnecessary; if it's deleted, yielding > main :: IO () > main = do n <- (getStdRandom (randomR (1,6))) :: IO Int > print n the code works just fine. Jon Cast

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