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In my attempts to grasp Existential Types in Haskell I decided to implement an integer-based fixed-length vector data type. I'm using ghc 7.8.3.

Specifically I wanted to write a program which asks the user for a possible new value to append to a fixed-length vector and then displays the resulting vector.

First I wrote a first version of the program like this:

{-# LANGUAGE GADTs                     #-}
{-# LANGUAGE ExistentialQuantification #-}
{-# LANGUAGE Rank2Types                #-}

import System.IO (hFlush, stdout)

data Z
data S n

data Vect n where
    ZV :: Vect Z
    CV :: Int -> Vect n -> Vect (S n)

data AnyVect = forall n. AnyVect (Vect n)

instance Show (Vect n) where
    show ZV = "Nil"
    show (CV x v) = show x ++ " : " ++ show v

vecAppend :: Int -> Vect n -> Vect (S n)
vecAppend x ZV = CV x ZV
vecAppend x v = CV x v

appendElem :: AnyVect -> IO AnyVect
appendElem (AnyVect v) = do
    putStr "> "
    hFlush stdout
    x <- readLn

    return $ if x == 0 then AnyVect v else AnyVect $ vecAppend x v

main = do
    AnyVect v <- appendElem $ AnyVect ZV
    putStrLn $ show v

Which works as expected. Then I decided to get rid of the unecessary AnyVect:

{-# LANGUAGE GADTs                     #-}
{-# LANGUAGE ExistentialQuantification #-}
{-# LANGUAGE Rank2Types                #-}

import System.IO (hFlush, stdout)

data Z
data S n

data Vect n where
    ZV :: Vect Z
    CV :: Int -> Vect n -> Vect (S n)

instance Show (Vect n) where
    show ZV = "Nil"
    show (CV x v) = show x ++ " : " ++ show v

vecAppend :: Int -> Vect n -> Vect (S n)
vecAppend x ZV = CV x ZV
vecAppend x v = CV x v

appendElem :: Vect n -> (forall n. Vect n -> a) -> IO a
appendElem v f = do
    putStr "> "
    hFlush stdout
    x <- readLn

    return $ if x == 0 then f v else f $ vecAppend x v

main = do
    appendElem ZV show >>= putStrLn

Which works as well, even if I don't really like the way main is written.

Is there any other simpler/cleaner way to write it?

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