- Beginners - Haskell.org

filterM function
by Shishir Srivastava 22 Apr '15

22 Apr '15

Hi, Continuing on from my previous question about powerset using filterM - Thanks to Alexey and Chaddaï. filterM is implemented as below - ---- filterM _ [] = return [] filterM p (x:xs) = do flg <- p x ys <- filterM p xs return (if flg then x:ys else ys) --- I still don't quite understand how 'flg' being a boolean [] is used in the last 'if statement' of implementation because when I try to do the same thing outside in GHCi it fails miserably even though I am casting it to [Int] - -- return (if [True, False] then "4" else "3")::[Int] -- Offtopic : Also if someone can explain how do I reply to individual mails/responses to my queries because I only get a digest on my gmail and there is no way to isolate and reply to individual mails either this list's page or from digest. Thanks, Shishir

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Powerset function
by Shishir Srivastava 21 Apr '15

21 Apr '15

Hi, In the 'learnyouahaskell' online book the powerset function is described like this - ---- powerset :: [a] -> [[a]] powerset xs = filterM (\x -> [True, False]) xs ---- And the filterM function is defined like this filterM :: (Monad m) => (a -> m Bool) -> [a] -> m [a] -- The filterM function is said to be an extension of 'filter' function which maps the monadic predicate over the given list. Now in powerset function the predicate returns a monad [True,False] for every element of the given list. So for e.g. if the input list was only [1] the output of filterM will understandably be the cartesian product of [True, False] x [1] = [[1], []]. Extending this if the given input list contains two elements [1,2] the predicate would be mapped one by one on each of the elements and the result combined which means the output should be [[1], [], [2], []] But the powerset of [1,2] is definitely not that. Please could someone help me in getting my head around with how filterM is working in this case. Thanks, Shishir

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What's the mean of >>=
by 김명신 19 Apr '15

19 Apr '15

It could be stupid question because i'm very beginner of haskell. I found strange literals when i read haskell code. Does anybody explan what the mean of >>= below ocode ? And additionally, What is mean of >> itself? putStrLn' [] = putChar '\n' putStrLn' xs - putStr' xs >>= \ x -> putChar '\n' putStr' [] = return () putStr' (x : xs) = putChar x >> putStr' n (because of i18n problem, back slash literal could be shown as '\') Thank you in advance Myung Shin Kim

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IO and ghci
by Mike Houghton 19 Apr '15

19 Apr '15

Hi, I have this in a .hs file and it works fine. That is I see lines printed to the screen when the .hs file is loaded and run from ghci. main = do file <- readFile "poem.txt" mapM_ putStrLn (lines file) Experimenting at the ghci command line with let f = readFile “poem.txt” the above is fine. But doing lines f gives <interactive>:52:7: Couldn't match type ‘IO String’ with ‘[Char]’ Expected type: String Actual type: IO String In the first argument of ‘lines’, namely ‘f’ In the expression: lines f In my simplistic beginners way I think this is because when I run main = do … everything is or is within an ‘impure’ area but at ghci command line it is not and I’m mixing pure/impure functions. So my questions are 1. Is this simplistic view any where near correct? 2. How can I progress at the command line in ghci? Many thanks Mike

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dropWhile returning last dropped element
by martin 18 Apr '15

18 Apr '15

Hello all Just curious: is there a library function like dropWhile, but which returns last dropped element along with the remainder of the list? If not, how can I write one - without explicit recursion - without traversing the list more than once - without traversing more of the list than necessary

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StateT MyState IO a vs ReaderT (IORef MyState) IO a
by Konstantin Saveljev 18 Apr '15

18 Apr '15

Hello, I'm working on a project where I use StateT MyState IO a to keep track of the state. MyState is deeply nested and I use lens to help me more easily access and modify the state. Just about yesterday I found someone mentioning about using ReaderT (IORef MyState) IO a instead of StateT MyState IO a because we are already in IO. Can someone explain to me what are the benefits of these different approaches? What about Garbage Collection in both cases (if there is any difference at all)? Can you use lens package to access and update the state which is hidden behind IORef ? Thanks, Konstantin Saveljev

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Re: [Haskell-beginners] StateT MyState IO a vs ReaderT (IORef MyState) IO a
by Grzegorz Milka 18 Apr '15

18 Apr '15

ReaderT (IORef MyState) IO a is introducing a global mutable state into your IO actions. This style is discouraged, because you have less guarantees about what can happen with your state. For example IORef is not protected from multi-threaded access. If you try to use this state in multiple threads you may run into bugs. StateT MyState IO a doesn't have that problem. For a typical use-case, that is an action that: reads state, performs action, modifies state, both are similar and neither should be significantly more comfortable. I would recommend to use StateT version unless you find that either it does not fit your model of computation or that it is underperforming. As for the performance it is hard to say. StateT allows the compiler to reason more about your code, so it might optimize better. However that's only one factor. You would need to test it to find out which one is better.

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Re: [Haskell-beginners] GCJ 2015 - Round 1A Problem - Haircut
by Doug McIlroy 18 Apr '15

18 Apr '15

> Can someone tell me how I could have avoided or fixed this? The trouble manifested as stack overflow on solving > https://code.google.com/codejam/contest/4224486/dashboard#s=p1 For efficiency, you'll probably need more cleverness in math than in Haskell. You can predict an approximate start time for the Nth customer's service from the average per-customer service time M, where 1/M = sum 1/M_k. Starting from that estimate, one can skip over almost the entire service simulation. Doug McIlroy

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GCJ 2015 - Round 1A Problem - Haircut
by Sourabh 18 Apr '15

18 Apr '15

So I decided to try the Google Code Jam this year, and decided to code in Haskell. I passed the qualifiers, and today was Round 1A. I managed to get the first problem right (both small and large input), and got the 2nd problem right (for the small input). However, for the large input, it ran out of stack space with the following error: Stack space overflow: current size 33632 bytes Use '+RTS -Ksize -RTS' to increase it. Can someone tell me how I could have avoided or fixed this? It is also very possible that I had an algorithmic issue (needed to optimize my code or have a different algorithm for the large input perhaps?). Here is the problem description: https://code.google.com/codejam/contest/4224486/dashboard#s=p1 And here is my code: https://github.com/cbrghostrider/Hacking/blob/master/codeJam/2015/haircut/h… I made sure to use foldr, which is supposedly more stack efficient I believe. I don't know if the list append (++) was causing issues. If anyone can spot any problems, or provide suggestions, I would really appreciate it! Even though the round is done, I'd like to optimize this to make it run within 8 minutes, as intended. Thanks in advance! -- SSJ

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State Monad
by Thomas Jakway 17 Apr '15

17 Apr '15

I'm having some trouble using the state monad with a random number generator. I'm trying to follow the LYAH example (http://learnyouahaskell.com/for-a-few-monads-more) but apparently the state monad API has changed, so I'm using Control.Monad.Trans.State.Lazy (state monad from the transformers package). Side question: is using transformers a good idea? I've heard some people complain about mtl's performance, others don't seem to care. I'm far too new to be able to judge anything (does it even make sense to compare transformers vs. mtl?) but if one has overtaken the other I'd rather use that. Here's what I currently have: import Control.Monad.Trans.State.Lazy type GeneratorState = State StdGen genThree :: Int -> GeneratorState Int genThree listMax = do --highest index in the list let listMin = 0 :: Int --lowest index in the list generator <- get let (generatedMin, state) = randomR (listMin, listMax) generator return generatedMin Although it typechecks it doesn't seem like an improvement over just using StdGen by itself. What I think I should have: genThree :: Int -> GeneratorState Int genThree listMax = do --highest index in the list let listMin = 0 :: Int --lowest index in the list generatedMin <- state randomR (listMin, listMax) return generatedMin * *I feel like I really botched this--what am I missing? * *

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