Question

I create my own data type, and try to implement functor method as follow:

data Hieu a = Hieu [a] deriving (Show, Read, Eq)

instance Functor Hieu where
        fmap f (Hieu [x]) = Hieu (f [x])

It’s very simple piece of code but it failed. Can you explain why?

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Thanks for all your responses. Now I understand that I apply functor only for one case. I tried to rewrite as follow, without using map

data Hieu a = Hieu [a] deriving (Show, Read, Eq)

consHieu :: a -> (Hieu a) -> (Hieu a)
consHieu x (Hieu xs) = Hieu (x:xs)

instance Functor Hieu where
    fmap f (Hieu (x:xs)) = consHieu (f x) (fmap f (Hieu xs))
    fmap f (Hieu []) = Hieu []

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Thanks for all your responses. Now I understand that I apply functor only for one case. I tried to rewrite as follow, without using map

data Hieu a = Hieu [a] deriving (Show, Read, Eq)

consHieu :: a -> (Hieu a) -> (Hieu a)

consHieu x (Hieu xs) = Hieu (x:xs)

instance Functor Hieu where

    fmap f (Hieu (x:xs)) = consHieu (f x) (fmap f (Hieu xs))
    fmap f (Hieu []) = Hieu []

Answer 1

In the expression f [x] you’re applying f to a list, but that’s not allowed since the type signature of fmap is (a -> b) -> Hieu a -> Hieu b. You’re not allowed to restrict f to [a] -> [b].

Perhaps you meant to write

instance Functor Hieu where
   fmap f (Hieu [x]) = Hieu [f x]

This would compile, but it would only work if the list has exactly one element. The normal way of making Hieu a functor would be to use map to apply the function to all of the elements like so:

instance Functor Hieu where
   fmap f (Hieu xs) = Hieu (map f xs)