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Why is there no "non-empty list" type in the Haskell base libraries?


Why is there no "non-empty list" type in the Haskell base libraries?

By : ScreenGems98
Date : November 17 2020, 11:52 AM
I hope this helps you . It is in base now since GHC 8.0: https://hackage.haskell.org/package/base-4.9.0.0/docs/Data-List-NonEmpty.html
code :


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Haskell: return the "list" result of a function as a "list of lists" without using an empty list &qu

Haskell: return the "list" result of a function as a "list of lists" without using an empty list &qu


By : Duodraco
Date : March 29 2020, 07:55 AM
this will help This one utilizes pattern matching to complete the task of not producing empty interleaving lists in a single traversal:
code :
splitWith :: Eq a => (a -> Bool) -> [a] -> [[a]]
splitWith f list = case splitWith' f list of
  []:result -> result
  result -> result
  where
    splitWith' _ [] = []
    splitWith' f (a:[]) = if f a then [] else [[a]]
    splitWith' f (a:b:tail) =
      let next = splitWith' f (b : tail)
      in if f a
        then if a == b
          then next
          else [] : next
        else case next of
          [] -> [[a]]
          nextHead:nextTail -> (a : nextHead) : nextTail
main = do
  print $ splitWith (== 'a') "miraaaakojajeja234"
  print $ splitWith (== 'a') "mirrraaaakkkojjjajeja234"
  print $ splitWith (== 'a') "aaabbbaaa"
["mir","koj","jej","234"]
["mirrr","kkkojjj","jej","234"]
["bbb"]
Haskell: Create a list of only certain "kind" of type?

Haskell: Create a list of only certain "kind" of type?


By : user2236617
Date : March 29 2020, 07:55 AM
it fixes the issue First of all, I would recommend not using record types with algebraic types, because you end up with partial accessor functions. For example, it is perfectly legal to have the code position (Individual (Person "John" "Doe" Male) True), but it will throw a runtime error. Instead, consider something more like
code :
data GovClient = GovClient {
    govName :: String
    } deriving Show

data CompanyClient = CompanyClient {
    companyName :: String,
    companyID :: Integer,        -- Also, don't overwrite existing names, `id` is built-in function
    companyContact :: String,
    companyPosition :: String
    } deriving Show

data IndividualClient = IndividualClient {
    indvFullName :: Person,
    indvOffers :: Bool
    } deriving Show
data Client
    = GovOrg GovClient
    | Company CompanyClient
    | Individual IndividualClient
    deriving (Show)
isIndividualClient :: Client -> Bool
isIndividualClient (Individual _) = True
isIndividualClient _ = False

listIndividuals :: [Client] -> [IndividualClient]
listIndividuals clients = filter isIndividualClient clients
listIndividuals = filter isIndividualClient
isIndividualClient :: Client -> Bool
isIndividualClient (Individual _ _) = True
isIndividualClient _ = False

listIndividuals :: [Client] -> [Client]
listIndividuals clients = filter isIndividualClient clients
Haskell, create "multi-type" list

Haskell, create "multi-type" list


By : Michael Mascolino
Date : March 29 2020, 07:55 AM
I wish this help you Is there a way to make haskell type this expression? , As joke, you can do like this:
code :
{-# LANGUAGE OverloadedStrings #-}

import Data.String                                                          

instance Num Bool where                                                     
    fromInteger 0 = False
    fromInteger _ = True

instance Fractional Bool where
    fromRational 0 = False
    fromRational _ = True

instance IsString Bool where
    fromString "" = False
    fromString _  = True

ls = [4, 3.2, True, "home"]
import Data.Dynamic

ls = [toDyn (4 :: Int), toDyn (3.2 :: Double), toDyn True, toDyn "hello"]
 {-# LANGUAGE ExistentialQuantification #-}

 data Any = forall a. {- here can be restrictions on `a` => -} Any a

 ls :: [Any]
 ls = [Any (4 :: Int), Any (3.2 :: Double), Any True, Any "home"] 
 data Variant = I Int | D Double | B Bool | S String

 ls :: [Variant]
 ls = [I 4, D 3.2, B true, S "home"]
Haskell - "Type error in application": "unification would give infinite type"

Haskell - "Type error in application": "unification would give infinite type"


By : Will
Date : March 29 2020, 07:55 AM
this will help First of all, types in Haskell start with capital letters. If you use identifiers starting with lower case letters in a type signature, they're interpreted as type variables. So your type char -> [char] -> int is the same as a -> [a] -> b, which is not a sensible type for your function. Your want Char -> [Char] -> Int.
Secondly [] is not a valid value of type Int. Fixing your type signature should produce an error message that tells you this in less ambiguous terms.
There is no implicit reference conversion from "Derive Type" to "Base type" in Generics

There is no implicit reference conversion from "Derive Type" to "Base type" in Generics


By : srinivasan
Date : March 29 2020, 07:55 AM
this one helps. Here is the code , You are violating your generic constraint:
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