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How do ambigious types compile with Aeson
by Sevcsik András 12 Sep '15

12 Sep '15

Hi list, > ghci> decode $ fromStrict $ encodeUtf8 $ "{\"a\" : \"b\"}" > Nothing > ghci> decode $ fromStrict $ encodeUtf8 $ "{\"a\" : \"b\"}" :: Maybe Object > Just (fromList [("a",String "b")]) The question is: how can the first example compile? I would expect that since GHC cannot infer the type I want from Aeson to decode to, it would die with an error that "t0 type variable is ambigious". But instead, it compiles just fine, and fails runtime. How it's decided what type should Aeson try to decode to? What type is that gives me Nothing on whatever JSON input I give? Cheers, Andras Sevcsik -- Minden jót, Sevcsik András

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wierd quickcheck outcome
by Roelof Wobben 10 Sep '15

10 Sep '15

Hello, I have this function : fourDifferent:: Integer -> Integer -> Integer -> Integer -> Bool fourDifferent a b c d = ( a == b ) && ( a == c ) && (a == d) which I wants to test with this function : prop_fourDifferent :: Integer -> Integer -> Integer -> Integer -> Bool prop_fourDifferent a b c d = fourDifferent a b c d == ( a == b ) && ( a == c ) && (a == d) so I do quickCheck propFourDifferent and see this outcome : *Solution> quickCheck prop_fourDifferent *** Failed! Falsifiable (after 2 tests and 2 shrinks): 0 0 0 1 *Solution> fourDifferent 0 0 0 1 False *Solution> let a = 0 *Solution> let b = 0 *Solution> let c = 0 *Solution> let d = 1 *Solution> (a == b) && ( a == c) && ( a == d ) False *Solution> So why is it failing. both gives false , Roelof

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Applicative on Tree
by Mike Houghton 08 Sep '15

08 Sep '15

Hi, I’m looking at data Tree a = Node a [Tree a] deriving (Show) and trying to write the Applicative instance. I have I think a correct Functor : instance Functor (Tree) where fmap f (Node x tr) = Node (f x) ( fmap (fmap f) tr) but I’m well and truly stuck on Applicative :) I have instance Applicative (Tree) where pure x = Node x [] (Node f []) <*> tr = fmap f tr (Node f fs) <*> tr@(Node x xs) = ????? I cant get rid of the question marks! :) any pointers would be really appreciated! Thanks

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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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