- Beginners - Haskell.org

runtime error <<loop>> when using -O compile option
by Gerold Meisinger 29 Dec '10

29 Dec '10

Hello! I'm working on a computer game using Yampa and I get the following runtime error: $ myprog: <<loop>> when compiling with $ ghc --make MyProg.hs -o myprog -O (without -O it works fine) I stripped the bug down to the program below. What's funny is that the error disappears under certain "odd circumstances" (marked as #1-#4). My questions are: 1. How can I avoid this bug without introducing one of the "odd circumstances"? 2. Why is it that I get this error? 3. How would you hunt down such a bug? Originally I got no clue where it came from, so I just took the program apart piece by piece. {-# LANGUAGE Arrows #-} module Main (main) where import FRP.Yampa type ObjIn = Event () -- loop #1 --type ObjIn = Bool -- no loop #1 type ObjOut = (String, Int) -- loop #2 --type ObjOut = Int -- no loop #2 type GameObj = SF ObjIn ObjOut testObj :: GameObj testObj = proc hit -> do returnA -< ("testObj", 1) -- loop #2 -- returnA -< 1 -- no loop #2 process :: [GameObj] -> SF () [ObjOut] process objs = proc _ -> do rec gamestate <- par logic objs -< gamestate -- loop #3 (recursive definition!) -- -< [] -- no loop #3 returnA -< gamestate logic :: [ObjOut] -> [sf] -> [(ObjIn, sf)] logic gamestate objs = map route objs where route obj = (if null (foo gamestate) then NoEvent else NoEvent, obj) -- loop #1 -- (if null (foo gamestate) then False else False, obj) -- no loop #1 foo :: [ObjOut] -> [ObjOut] foo [] = [] foo objs = concat (collisions objs) where collisions [] = [] collisions (out:objs') = [[out, out'] | out' <- objs, out `collide` out'] -- loop #4 -- [[out, out'] | out' <- objs, True] -- no loop #4 collide :: ObjOut -> ObjOut -> Bool collide (_, p) (_, p') = True -- loop #2 --collide p p' = True -- no loop #2 main :: IO () main = do putStrLn . show $ embed (process [testObj]) ((), [(1.0, Nothing)]) (Btw: I re-opened a bug report: http://hackage.haskell.org/trac/ghc/ticket/2722#comment:10 )

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strange behaviour : computing lowest divisor
by Abhijit Ray 29 Dec '10

29 Dec '10

HI, I have written(copied) a small piece of code which finds the lowest divisor of am integer (greater than 1) the code is here ld::Int->Int ld n = ld' 2 n where ld' i n | rem n i == 0 = i | i^2 > n = n | otherwise = ld' (i+1) n now it seems to work fine for small numbers but for a big number we get *Main> ld 278970415063349480483707695 7 the number is obviously divisible by 5 , so where is the anomaly ? Thanks, Abhijit On Wed, Dec 29, 2010 at 2:52 PM, Alex Rozenshteyn <rpglover64(a)gmail.com>wrote: > I'm trying to understand the technique referred to as "tying the knot", but > documentation on the internet seems to be much sparser and more obtuse than > I like. > > So I'm asking here. > > As far as I understand, "tying the knot" refers to a way of using laziness > to implement something like references in a purely functional way. > > I'm trying to write a toy simulation: > I have a population :: [Person] > I want to collect a random subset of possible pairs of distinct people. > So I go to each person in the population and select a subset of the people > after him/her in the list; these are pairs in which s/he is the first > element. > > I want to then be able to ask for all pairs in which a person is the first > or the second element. I could give each person a unique id, but it seems > like tying the knot is a valid way to implement this. > > Please correct me if I am misunderstanding. > > Thank you. > > -- > Alex R > > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners > >

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lambda
by John Moore 28 Dec '10

28 Dec '10

Hi, Can someone explain how this lambda is suppose to work, I cannot figure out what it does? data Lambda = Val Double | Add Lambda Lambda | Subtract Lambda Lambda | Multiply Lambda Lambda | Divide Lambda Lambda | Var String | Let String Lambda Lambda deriving Show demo1 = (Add(Multiply(Divide(Subtract(Val 25)(Val 5))(Val 10))(Val 7))(Val 30)) --let x = 1+1 in 3 - x test1 = Let "x" (Add (Val 1) (Val 1)) (Add(Var "x")(Var "x")) test6 = Let "y" (Add (Val 7)(Val 6)) (Subtract (Val 6)(Var "y")) -- 4 * (let x = 1+1 in 3 + x) test2 = Multiply (Val 4) test1 -- x * (let x = 1+1 in 3 + x) test3 = Let "y" (Add (Val 4)(Val 7))(Multiply test1 (Var "y")) test5 = Add (test1) test6 type Dict =[(String,Lambda)]-- dictionary which takes a string and an expression emptyDict :: Dict emptyDict = []-- empty dictionary null set addEntry :: String->Lambda ->Dict -> Dict addEntry= \n e d -> (n,e): d -- add an entry we take a string,expression and a dictionary :gives back dictionary lookupEntry :: String -> Dict -> Maybe Lambda lookupEntry n [] = Nothing lookupEntry n (x:xs) = if (n == k)-- if the entry is equal to then (Just v) else lookupEntry n xs -- find the value of n in the rest of the list where (k,v) = x -- evalStep :: Dict -> Lambda -> Lambda evalStep d (Val x) = (Val x)-- we have a dictionary and a value return the value evalStep d(Add x y)-- dictionary and a add expression of x y = case x of (Val a) -> case y of (Val b) -> Val (a+b) left -> Add x (evalStep d y) right -> Add (evalStep d x)y evalStep d(Subtract x y) = case x of (Val a) -> case y of (Val b) -> Val (a-b) left -> Subtract x (evalStep d y) right -> Subtract (evalStep d x)y evalStep d(Multiply x y) = case x of (Val a) -> case y of (Val b) -> Val (a*b) left -> Multiply x (evalStep d y) right -> Multiply (evalStep d x)y evalStep d (Divide x y) = case x of (Val a) -> case y of (Val b) -> Val (a/b) left -> Divide x (evalStep d y) right -> Divide (evalStep d x)y evalStep d (Let n e1 e2) = case e1 of (Val a) -> case e2 of (Val b)-> Val b left -> evalStep (addEntry n e1 d)e2 right -> evalStep (addEntry n e1 d) e2 evalStep d (Var x) = case lookup x d of Just e -> e Nothing -> error "Error in expression -- no definition for variable!" evaluate :: Dict-> [Lambda] -> Lambda -> IO() evaluate d(x:xs) e = do putStrLn (show e) putStrLn "Do another step (y/n) or rollback (r)? :" c <- getLine case c of "y" -> let e'= (evalStep d e)in evaluate d (e:x:xs) e'-- build up history "r" -> case (x:xs) of (x:xs)-> evaluate d xs x []-> do { putStrLn "Empty" ;evaluate d(x:xs) e } "n" -> putStrLn $ "Ok you said no :" ++ c ev :: Dict-> [Lambda] -> Lambda -> IO() ev = \d (x:xs) e ->do putStrLn (show e) putStrLn "Do another step (y/n) or rollback (r)? :" c <- getLine case c of "y" -> let e'= (evalStep d e)in evaluate d (e:x:xs) e'-- build up history "r" -> case (x:xs) of (x:xs)-> evaluate d xs x []-> do { putStrLn "Empty" ;evaluate d(x:xs) e } "n" -> putStrLn $ "Ok you said no :" ++ c isFree :: String -> Lambda -> Bool isFree n1 (Let n2 e1 e2) |n1 == n2 = False |otherwise = True subst :: String -> Lambda -> Lambda->Lambda subst n e1 e2 |isFree n e2 = subst' n e1 e2 |otherwise = e2 subst' :: String -> Lambda -> Lambda->Lambda subst' n0 e1 (Var n1) |n0 == n1 = e1 |otherwise = Var n1 subst' s v (Add e1 e2) = Add(subst' s v e1)(subst' s v e2) subst' s v (Multiply e1 e2) = Multiply(subst' s v e1)(subst' s v e2) subst' s v (Divide e1 e2) = Divide(subst' s v e1)(subst' s v e2) subst' s v (Subtract e1 e2) = Subtract(subst' s v e1)(subst' s v e2) subst' s v (Let n e1 e2) = Let n (subst' s v e1)(subst' s v e2) John

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Infix operator and precedence
by Patrick LeBoutillier 24 Dec '10

24 Dec '10

Hi, I'm just messing around with infix operators, and I'm trying to implement some kind of ternary if operator: (??) :: Bool -> (Bool -> a) -> a (??) = flip ($) (?:) :: a -> a -> (Bool -> a) t ?: f = \b -> if b then t else f test b = b ?? ("IF" ?: "THEN") I would like to get rid of the parentheses around the ("IF" ?: "THEN") part, but I'm not familiar with the use of theinfixr/infixl functions. I'm not sure which one to use in this case and how exactly to use it. Can anyone shed some light? Thanks a lot, Patrick -- ===================== Patrick LeBoutillier Rosemère, Québec, Canada

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how to input a Let expression
by John Moore 22 Dec '10

22 Dec '10

Hi, I'm missing something basic here, I wrote this program a while back. But I cant remember how to work it. Would someone please explain how I enter a value for a variable x,y etc and get this to work. It goes by entering evaluate [][test1]test1 but I cant remember how to write in The Let part and not have it keep ending in -no definotion of a variable. John

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How to think in Haskell
by Jun HU 21 Dec '10

21 Dec '10

Dear everyone, My first language is C, and I've strong intention in learning a pure functional programming language. My very first question is how to think in the functional programming way, anyone has some ideas. Really appreciate.... Regards, Jun

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Forum vs mailing list
by Amy de Buitléir 21 Dec '10

21 Dec '10

On 21 December 2010 12:42, David Jameson <dhjdhj(a)gmail.com> wrote: > P.S. By the way, is is not possible to have a forum instead of a mailing list --- these mailing lists are so 19th century! If you use an face like GMANE, then you'll get the best of both worlds, plus a bit more. Youll be able to post messages using a forum-like interface. And you don't have to be subscribed to the mailing list to post through GMANE, so if you don't want your inbox filling up, simply unsubscribe. You'll still be able to read all the other posts, but you'll have to go to the GMANE page to do so. Look here for more information: http://dir.gmane.org/gmane.comp.lang.haskell.beginners

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Uncalled for apology.
by Sean Charles 21 Dec '10

21 Dec '10

I felt compelled to PM Paul about his email. In support. If this list is to be vetted and thought-controlled by self-imposed politically correct Ministry of Thought operatives then I am leaving today. It's a shame that the mindset required to understand Haskell is still open to petty thinking and the desire to impose its thoughts upon others. Freedom of speech, thought and expression. Well, that's ruined my Christmas I think. What is it about human beings that makes them unable to accept what somebody else has to say without wanting to pontificate ? Sigh, and it was such a nice list too... :(

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Standard Haskell Module Files
by Jane Ren 21 Dec '10

21 Dec '10

Hi, Could someone tell me where I can get the standard Haskell module files such as source code for the Data module? Thanks

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Equivalent of inheritance in Haskell
by C K Kashyap 21 Dec '10

21 Dec '10

Hi, In oops, one can do reuse code by extending a parent class. What is the equivalent in Haskell? Regards, Kashyap

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