Haskell Type issue

Question

I wrote the code:

{-# LANGUAGE InstanceSigs #-} {-# LANGUAGE FlexibleInstances #-} module MonoidApp where class Monoid' a where mempty' :: a mappend' :: a -> a -> a mconcat' :: [a] -> a mconcat' = foldr mappend' mempty' instance Monoid' Int where mempty' :: Int a => a mempty' = 0 mappend' :: Int a => a -> a -> a mappend' a b = (+) a b

But it runs into error:

‘Int’ is applied to too many type arguments In the type signature for ‘mempty'’: mempty' :: Int a => a In the instance declaration for ‘Monoid' Int’ Failed, modules loaded: none.

Any ideas why?

The error message is pretty clear: Int is a type with no parameters, but you've tried to apply it to one. I'm guessing you actually mean mempty' :: Int; mappend' :: Int -> Int -> Int. — Benjamin Hodgson
– Benjamin Hodgson, Commented Mar 28, 2016 at 22:15
What if you leave out lines mempty' :: Int a => a and mappend' :: Int a => a -> a -> a? — J.J. Hakala
– J.J. Hakala, Commented Mar 28, 2016 at 22:17
@J.J.Hakala Then I cannot use the function with parameter (mappend' 1 2): No instance for (Num a0) arising from a use of ‘it’ — barbara
– barbara, Commented Mar 28, 2016 at 22:26
I need to a type every time to make it work: mappend' (1::Int) (2::Int) — barbara
– barbara, Commented Mar 28, 2016 at 22:35
Don't change your question in a way that doesn't include your original problem. If you have additional information add an update section, if you have new questions open a separate question. — epsilonhalbe
– epsilonhalbe, Commented Mar 29, 2016 at 0:17

Bolo · Accepted Answer · 2016-03-28 22:17:25Z

3

It should be:

instance Monoid' Int where mempty' :: Int mempty' = 0 mappend' :: Int -> Int -> Int mappend' a b = (+) a b

answered Mar 28, 2016 at 22:17

Bolo

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