CODEWITHSUNDEEP
parsinghaskell
digitalguy99
digitalguy99

Reputation: 549

Haskell Date Parsing with Custom Separator

I'm trying to print out something like this:

enter image description here

Here's my code:

import System.Environment
import Data.Time

main = do args <- getArgs
          let year = read $ args !! 0 
          let month = read $ args !! 1 
          let day = read $ args !! 2 
          let greg = fromGregorian year month day
          print $ showDateFormat $ toGregorian $ addDays 10 $ greg
          print $ showDateFormat $ toGregorian $ addDays 100 $ greg
          print $ showDateFormat $ toGregorian $ addDays 1000 $ greg
          print $ showDateFormat $ toGregorian $ addDays 10000 $ greg

showDateFormat :: (Integer,Int,Int) -> String
showDateFormat (y,m,d) = y ++ "/" ++ m ++ "/" ++ d ++ "\n"

I can't figure out what's wrong.

This is the error I got:

enter image description here

Upvotes: 0

Views: 72

Answers (1)

chepner
chepner

Reputation: 532238

Haskell will not implicitly convert a value of one type to another; you have to do such things explicitly. In this case, you can use show to convert an Int or an Integer to a string containing its base-10 representation.

showDateFormat :: (Integer,Int,Int) -> String
showDateFormat (y,m,d) = show y ++ "/" ++ show m ++ "/" ++ show d ++ "\n"

The error message is literally telling you that it expects y to be a value of type [Char] (aka String), because ++ (receiving "/" :: String as one argument) expects a String as the other, but you have pass an Integer value for y instead.

(Note that String is the expected type because it is a valid type for ++, while Integer is not. Otherwise, the type checker would use the left-hand argument to fix the type. [1::Int] ++ "foo", for example, fails because "foo" does not have type [Int], not because [1] does not have type [Char].)

Upvotes: 1

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