June 2012 - Beginners

Understanding how laziness works
by Matt Ford 19 Jun '12

19 Jun '12

Apologies for the maybe double post (no subject and hadn't confirmed the mailing list first time I sent this). Hi, Here's a simple expression double ( double [1,2,3,4] ) !! 0 The 1st element of a list is double ( 1*2 : double [2,3,4] ) !! 0 1*2*2 : double ( double [2,3,4 ] ) !! 0 1*2*2 4 calculated without having to traverse the rest of the list due to laziness. But how far does the laziness extend? For example, if I take the 2nd element double (double [1,2,3,4]) !! 1 Do we have 1*2*2 : double ( 2*2 : double [3,4] ) !! 1 1*2*2 : 2*2*2 : double ( double [3,4] ) !! 1 2*2*2 8 Resulting in 8. I assume that the 1st list entry 1*2*2 doesn't actually get calculated, but the expression does get formed? The second element (index 1) gets formed, isolated (by !!) and calculated as the return value. And then things can happily stop. What I'm checking is that the laziness doesn't some how magically extend to not having to form the preceding expressions that make up the resulting list? Or can Haskell recognize that this is like doing function composition i.e., (double . double) [ 1 2, 3, 4 ] !! 1 Thanks for any answer. Cheers, Matt. p.s, I kinda hope things are how I've assumed as the example finally shows me some concrete benefit of function composition which has never seemed important given it's so easily re-written as application.

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problem with type
by miro 17 Jun '12

17 Jun '12

Am a bit stuck here,... please, what is wrong with this? checkNode :: String -> [String] -> Bool checkNode s nodes = [s == node | node <- nodes ] src/me.hs:3:4: Couldn't match expected type `Bool' with actual type `[t0]' [1 of 1] Compiling Main ( src/me.hs, interpreted ) In the expression: [s == node | node <- nodes] Failed, modules loaded: none. In an equation for `checkNode': checkNode s nodes = [s == node | node <- nodes] thanks, Miro

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how monomorphism restriction effects eta reduction
by Song Zhang 17 Jun '12

17 Jun '12

I think monomorphism restriction is for shared evaluation.like following function f xs = let len = genericLength xs in (len, len) from haskell wiki under monomorphism the type is Num t => [b] -> (t, t), if it no monomorphism restriction the type will be (Num t, Num t1) => [b] -> (t, t1) which lead to twice evaluation of len. However I didn't see how the restriction effects eta reduction, when define function like f = show, g = (==) or f = \x -> show x, g = \x -> \y -> (x==y). anyone can explain it to me? Thanks a lot.

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lambda notation vs function-argument notation in GHCi
by Morel Pisum 17 Jun '12

17 Jun '12

Let's define a function in GHCi: > Prelude> let f s n = and (map (==s) n) > Prelude> :t f > f :: Eq a => a -> [a] -> Bool This is fine. But when I define this function using lambda notation, I get this > Prelude> let f = \s n -> and (map (==s) n) > Prelude> :t f > f :: () -> [()] -> Bool which is really weird. Why does this happen?

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Simplify (normalize) symbolic polynom-like expression
by Daniel Hlynskyi 17 Jun '12

17 Jun '12

Hello. I am new to typefull programming, so I've got a question. I have a simple mathematical expression (addition, product and exponentiation only): > data Expr = I Int -- integer constant > | V -- symbolic variable > | Sum [Expr] > | Prod [Expr] > | Pow Expr Expr What I want is normalize\simplify this expression. Eg `Prod [Pow V (I 0), Pow V (I 1)] ` must be simplified to just `V`. What techniques should I use to write `normalize` function? Simplification rules are quite simple: > normalize (Sum [a]) = normalize a > normalize (Sum xs) | (I 0) `elem` xs = map nomalize . Sum $ filter (/= I 0) xs > | otherwise = map normalize xs > normalize (Prod xs) | (I 0) `elem` xs = I 0 > normalize (Prod xs) | (I 1) `elem` xs = map nomalize . Prod $ filter (/= I 1) xs > | otherwise = map normalize xs > normalize (Pow a (I 0)) = I 1 > normalize (Pow a (I 1)) = normalize a and so on. But rules like theese cannot simplify some expressions, for example, `Prod [Pow V (I 0), Pow V (I 1)] `.

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IO - getContents - putStrLn
by Robert Heumüller 16 Jun '12

16 Jun '12

Hi, this is probably an easy question, but i simply can't figure out, why this does not work: import Data.Char main = do contents <- getContents putStrLn $ show $ splitcomma contents splitcomma = split ',' split :: Char -> String -> [String] split _ "" = [""] split sp (c:cs) | c == sp = "": rest | otherwise = (c : head rest) : tail rest where rest = split sp cs The program compiles and runs without any problems. But there is absolutely no output, when f.eg. i type "hello, world" and hit return. Why would that be? Thank you very much

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about kind of (->)
by Song Zhang 16 Jun '12

16 Jun '12

a function has also a kind. According to haskell report 2010 4.1.2 it is * -> * -> *, which is easy to understand. However in ghci I type :k (->). the output is ?? -> ? -> *. I want to know what do ?? and ? mean. Thanks

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the "download" package fail to install
by Song Zhang 15 Jun '12

15 Jun '12

Hi There is a package for High-level file download based on URLs called "download", but failed to build on windows, If you don't know it, you can refer to hackage first. I compiled in C, but did not get useful massage, anyone can help me a little to fix that. cabal install download get a bad header file

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Difference between Char and Data.Char modules
by Ken Overton 14 Jun '12

14 Jun '12

I just started looking at the Happy packag; I get a compile error when attempting to build examples/glr/expr-eval: ExprData.hs:9:8: Could not find module `Char' It is a member of the hidden package `haskell98-2.0.0.1'. ExprData.hs:9 says “import Char” and if I change this to “import Data.Char” it compiles correctly (then I get a similar message for System, but when I understand Char I expect I will understand this too). Could someone kindly explain the difference between these two modules? From the look of the error, I’m guessing there’s a history lesson involved? Thanks, Ken ________________________________ This email and any attachments may contain information which is confidential and/or privileged. The information is intended exclusively for the addressee and the views expressed may not be official policy, but the personal views of the originator. If you are not the intended recipient, be aware that any disclosure, copying, distribution or use of the contents is prohibited. If you have received this email and any file transmitted with it in error, please notify the sender by telephone or return email immediately and delete the material from your computer. Internet communications are not secure and Lab49 is not responsible for their abuse by third parties, nor for any alteration or corruption in transmission, nor for any damage or loss caused by any virus or other defect. Lab49 accepts no liability or responsibility arising out of or in any way connected to this email.

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HXT and missing or null values
by Ian Knopke 14 Jun '12

14 Jun '12

I got it. For anyone else running into the problem from my earlier email, the way to get a null string is as follows: Replace this line: second <- text <<< atTag "second" -< x with this: second <- ((text `orElse` (constA "")) <<< atTag "second" -< x Fixed program: data Thing = Thing {first :: String, second :: String} getThings = atTag "thing" >>> proc x -> do first <- text <<< atTag "first" -< x second <- ((text `orElse` (constA "")) <<< atTag "second" -< x returnA -< Thing {first = first,second = second} atTag tag = deep (isElem >>> hasName tag) text = getChildren >>> getText parseThings str = runX (readString [withValidate no] str >>> getThings)

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