Defining a new monad in haskell raises no instance for Applicative

HaskellMonads

Haskell Problem Overview


I am trying to define a new monad and I am getting a strange error

newmonad.hs

newtype Wrapped a = Wrap {unwrap :: a}
instance Monad Wrapped where
(>>=) (Wrap x) f =  f x
return x = Wrap x

main = do putStrLn "yay"

$ ghc --version
The Glorious Glasgow Haskell Compilation System, version 7.10.1

$ ghc newmonad.hs [1 of 1] Compiling Main ( newmonad.hs, newmonad.o )

newmonad.hs:2:10: No instance for (Applicative Wrapped) arising from the superclasses of an instance declaration In the instance declaration for ‘Monad Wrapped’

Why do I need to define an instance of Applicative?

Haskell Solutions


Solution 1 - Haskell

This is the Applicative Monad Proposal (AMP). Now whenever you declare something as Monad, you also have to declare it as Applicative (and therefore Functor). Mathematically speaking, every monad is an applicative functor, so this makes sense.

You can do the following to remove the error:

instance Functor Wrap where
  fmap f (Wrap x) = Wrap (f x)

instance Applicative Wrap where
  pure = Wrap
  Wrap f <*> Wrap x = Wrap (f x)

https://wiki.haskell.org/Functor-Applicative-Monad_Proposal

Edit: Maybe I should point out more clearly that this is a recent thing? The code you posted used to work before, but with recent versions of GHC you'll get an error. It's a breaking change.

Edit: The following declarations should work for any monad:

import Control.Applicative -- Otherwise you can't do the Applicative instance.
import Control.Monad (liftM, ap)

instance Functor ??? where
  fmap = liftM

instance Applicative ??? where
  pure  = return
  (<*>) = ap

Depending on the monad in question, there may be more efficient implementations possible, but this is a simple starting point.

Solution 2 - Haskell

The most normalized and unobtrusive answer is :-

as Monad is dependent upon Applicative

> class Applicative m => Monad m where ...

and Applicative is dependent upon Functor

> class Functor f => Applicative f where ...

we need the instance definitions

> instance Functor Wrapped where
>     fmap = liftM

and

> instance Applicative Wrapped where
>     pure = return
>     (<*>) = ap

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionfakedrakeView Question on Stackoverflow
Solution 1 - HaskellMathematicalOrchidView Answer on Stackoverflow
Solution 2 - HaskellAaronNGrayView Answer on Stackoverflow