August 2015 - Beginners

Multiline code in GHCi
by Shishir Srivastava 29 Aug '17

29 Aug '17

Hi, I am trying to run this basic multiline code in GHCi. I am not sure if this is even possible without writing it in a file and compiling it in a traditional way. ------- Prelude> :set +m Prelude> let main1 = do Prelude| a <- readLn Prelude| return a Prelude| Prelude> Prelude> main1 <interactive>:75:1: No instance for (Show (IO b0)) arising from a use of `print' In the first argument of `print', namely `it' In a stmt of an interactive GHCi command: print it ------- Could someone please confirm if the function can be defined and called like this in GHCi or point to where am going wrong. Thanks, Shishir Srivastava

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applicative style
by Williams, Wes(AWF) 24 Sep '15

24 Sep '15

Hi haskellers, I am trying to understand why I get the following error in learning applicative style. Prelude> let estimates = [5,5,8,8,2,1,5,2] Prelude> (/) <$> Just $ foldl (+) 0 estimates <*> Just . fromIntegral $ length estimates <interactive>:54:1: Non type-variable argument in the constraint: Fractional (Maybe r) (Use FlexibleContexts to permit this) When checking that 'it' has the inferred type it :: forall a r. (Fractional (Maybe r), Num a, Num (Int -> Maybe a -> r)) => Maybe r -> Maybe r All the parts work individually. If use let and assign the parts to x and y it also works. E.g. This works let x = Just $ foldl (+) estimates Let y = Just . fromIntegral $ length estimates (/) <$> x <*> y I clearly do not understand exactly how these work. :-) Thanks for any help, -wes

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Config data
by Mike Houghton 18 Sep '15

18 Sep '15

Hi, I’ve haskell files parser.hs and email.hs email.hs is a module imported into parser and parser has the main. email has various passwords and server names needed to connect and send email. I want to have those details in a config file and read it in at start - using say config module. I’m ok with this part but the practicality of doing it eludes me... It seems to me that email module can either get the config details itself or be told them by parser. If email wants to get them how would it do this? It does not have a main that gets run so can’t load a config file.(can it ??) However parser can load the config in its main and then tell mail what the values are but how would mail save them? Of course I could chanage the signatures of the email send/receive functions in mail to take the connection details but that seems wrong. Thanks Mike

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Stack and packages not in the snapshot
by Thomas Koster 31 Aug '15

31 Aug '15

Good morning list, I have started to use stack with Stackage LTS for building my projects because I like the idea of a stable snapshot of packages known to work together. Then I tried to build "x509-util-1.5.2". This package fails to build and this is already known by upstream. And lo, "x509-util" is not in Stackage LTS 3.2, which makes sense. But what surprised me is that stack attempted to build a package not in the snapshot, when my resolver is "lts-3.2" (and no extra-deps). Is there a way to configure stack to build packages from the snapshot only? Or does it already work this way, only trying "x509-util" because *I* said "stack build x509-util"? Thanks in advance. -- Thomas Koster

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Difficulty understanding the Towers of Hanoi exercise from CIS194
by John Del Rosario 30 Aug '15

30 Aug '15

I've just completed the first exercise (Credit Card Number Validation) and was quite happy with myself. But moving on to the next exercise I found myself having trouble trying to understand some parts of the question. I've solved Towers of Hanoi before with an imperative language, but I'm having trouble visualising the steps described in this paragraph: To move n discs (stacked in increasing size) from peg a to peg b using peg c as temporary storage, 1. move n-1 discs from a to c using b as temporary storage 2. move the top disc from a to b 3. move n-1 discs from c to b using a as temporary storage. I've never seen the solution described this way before. So if I start with 5 discs, then for the first step I'll be moving 4 (n-1) discs? But can't I only move 1 disc at a time? And step 2 says to move the top disc from a to b. Which disc is the top disc at this point? I'm still in a very imperative mindset (and having no formal training doesn't help), but I'm hoping that will improve as I go through this course!

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recursion - more elegant solution
by Miro Karpis 29 Aug '15

29 Aug '15

Hi haskellers, I have made one recursive function, where input is: nodesIds: [1,2,4,3,5] edgesIds: [(3,5),(4,3),(2,4),(1,2)] and on output I need/have: [([1,2,4,3,5],(3,5)),([1,2,4,3],(4,3)),([1,2,4],(2,4)),([1,2],(1,2))] I have made one function for this, but it seems to me a bit too big and 'ugly'? Please, do you have any hints for a more elegant solution? joinEdgeNodes' :: [Int] -> [Int] -> [(Int, Int)] -> [([Int], (Int, Int))] joinEdgeNodes' [] _ [] = [] joinEdgeNodes' [] _ _ = [] joinEdgeNodes' _ _ [] = [] joinEdgeNodes' (n) (wn) [e] = [(n, e)] joinEdgeNodes' (n) [] (e:es) = joinEdgeNodes' n edgeNodes es ++ [(edgeNodes, e)] where edgeNodes = take 2 n joinEdgeNodes' (n) (wn) (e:es) = joinEdgeNodes' n edgeNodes es ++ [(edgeNodes, e)] where edgeNodes = take edgeNodesLength n edgeNodesLength = (length wn) + 1 I call the function with: let l = joinEdgeNodes' nodeIds [] edgeIds Cheers, Miro

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applicative style
by Williams, Wes(AWF) 28 Aug '15

28 Aug '15

Awesome, I clearly am off on my understanding. How could I do this: "foldl (+) 0 estimates <+> (Just . fromIntegral) (length estimates)" without the parenthesis? Thanks, Wes

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converting a json encoded radix tree to a haskell data type
by Adam Flott 28 Aug '15

28 Aug '15

I'm having trouble converting a JSON encoded Radix tree into a Haskell data type[1]. I've tried numerous ways to get the FromJSON instances to handle all cases, but failing miserably. [1] unfortunately these type layouts are unchangeable as they are auto generated types from Thrift Here is a stripped down version of what I'm working with. For the JSON file, all keys are unknown at parse time. I only know that a key will be a string and it's value will be another JSON object or a JSON array of fixed length. Any help is appreciated. -- radix.hs -- {-# LANGUAGE FlexibleInstances #-} {-# LANGUAGE OverloadedStrings #-} import Control.Monad (mzero) import Data.Int import Data.Typeable import Data.Aeson import qualified Data.ByteString.Lazy as BL import Data.Text.Lazy as TL import qualified Data.Vector as V data Things = MkThings { thing :: TL.Text, times :: ThingTimes } deriving (Show, Eq, Typeable) data ThingTimes = MkThingtimes { ml :: V.Vector Times } deriving (Show, Eq, Typeable) data Times = MkTimes { t1 :: Maybe Int32, t2 :: Maybe Int32 } deriving (Show, Eq, Typeable) instance FromJSON (V.Vector Things) where parseJSON _ = return V.empty decodeRadix ::BL.ByteString -> Either String (V.Vector Things) decodeRadix = eitherDecode main :: IO () main = do j <- BL.readFile "radix.json" case decodeRadix j of Left err -> error err Right r -> print r -- radix.hs -- -- radix.json -- { "a" : { "b" : [ 1, 2 ], "c" : { "d" : [ 3, null ] } }, "a2" : { "b2" : [ 4, 5 ] } } -- radix.json --

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Websockets Library
by Noah Sluss 23 Aug '15

23 Aug '15

Hello, I've been trying to use the WebSockets library to connect to the coinbase feed. The library seems to be doing something behind the scenes that I'm not expecting. Here is my code: > {-# LANGUAGE ViewPatterns #-} > {-# LANGUAGE OverloadedStrings #-} > > import Data.Text > import Data.Text.Encoding (decodeUtf8) > import Network.WebSockets > import qualified Data.ByteString.Lazy as LBS > > main :: IO () > main = runClient "ws-feed.exchange.coinbase.com" 8080 "/" > handleConnection > handleConnection connection = do > send connection initSub > let loop = do > priceMsg <- receiveDataMessage connection > print priceMsg > loop > > initSub :: Message > initSub = DataMessage $ Text "{\"type\":\"subscribe\", > \"product_id\":\"BTC-USD\"}" When I run this, I get a print out showing a malformed response exception with a Moved permanently message. Now, it also shows the location that I was trying to connect to, which is: "https://ws-feed.exchange.coinbase.com/ ". This isn't right, because the websocket feed uses the wss:// protocol rather than https://. i've tried changing the url to be "wss:// ws-feed.exchange.coinbase.com/", but when I try that, dns resolution fails. If I had to guess, I could imagine that it tries to look for "https://wss:// ws-feed.exchange.coinbase.com", which would obviously not work. I'm wondering if there is a way within the library to change the protocol that client uses. If there is, it doesn't seem to be documented. Any insights as to what's going on here would be incredibly helpful. Thanks, Noah

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Re: [Haskell-beginners] oddsFrom3 function
by Doug McIlroy 17 Aug '15

17 Aug '15

> > oddsFrom3 :: [Integer] > > oddsFrom3 = 3 : map (+2) oddsFrom3 > > > > > > Thanks for your help. > > Try to expand a few steps of the recursion by hand e.g.: > > 3 : (map (+2) (3 : map (+2) (3 : map (+2) ...))) > > > As you can see, the deeper you go more 'map (+2)' are applied to '3'. Some more ways to describe the program, which may be useful: As with any recursive function, assume you know the whole series and then confirm that by verifying the inductive step. In this case oddsFrom3 = [3,5,7,9,11,...] map (+2) oddsFrom3 = [5,7,9,11,13,...] voila oddsFrom3 = 3 : map (+2) oddsFrom3 Assuming we have the whole series, we see its tail is computed from the whole by adding 2 to each element. Notice that we don't actually have to know the values in the tail in order to write the formula for the tail. Yet another way to describe the program: the "output" is taken as "input". This works because the first element of the output, namely 3, is provided in advance. Each output element can then be computed before it is needed as input. In an imperative language this would be done so: integer oddsFrom3[0:HUGE] oddsFrom3[0] := 3 for i:=1 to HUGE do oddsFrom3[i] = oddsFrom3[i-1] + 2

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