Constructing Haskell data types with many fields

Question

I have data type

data Constants = Constants { x1 :: Int, x2 :: Int, ... , x100 :: Int }

Names x1, x2, ..., x100 may have irregular naming.

Is there an elegant way to create Constants object from list of 100 Int values?

mkConstants :: [Int] -> Constants
mkConstants [x1, x2, ..., x100] = Constants x1 x2 ... x100 -- The bad way

What about reverse task?

extractInts :: Constants -> [Int]
extractInts (Constants x1 x2 ... x100) = [x1, x2, ..., x100] -- The bad way

Answer 1

Edit: see Reading long data structure in Haskell for a modified version when the record fields are of different types (i.e. not all ints).

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One possibility would be to use type classes:

{-# LANGUAGE FlexibleInstances #-}

data Constants = Constants { a :: Int, b :: Int, c :: Int, d :: Int, e :: Int }
    deriving Show

class Cons a where
    cons :: a -> [Int] -> Maybe Constants

instance Cons Constants where
    cons c [] = Just c
    cons _ _  = Nothing

instance (Cons a) => Cons (Int -> a) where
    cons f (x:xs) = cons (f x) xs
    cons _ _      = Nothing

then, if the list is of the right size:

\> cons Constants [1..5]
Just (Constants {a = 1, b = 2, c = 3, d = 4, e = 5})

and otherwise you get nothing:

\> cons Constants [1..4]
Nothing
\> cons Constants [1..6]
Nothing