[Haskell-cafe] Re: Re: Monads from Functors

David Menendez wrote:

On Wed, Apr 8, 2009 at 5:20 PM, Ben Franksen wrote:

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Sebastian Fischer wrote:

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{-# LANGUAGE Rank2Types #-}

Dear Haskellers,

I just realized that we get instances of `Monad` from pointed functors and instances of `MonadPlus` from alternative functors.

Is this folklore?

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import Control.Monad import Control.Applicative

In fact, every unary type constructor gives rise to a monad by the continuation monad transformer.

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newtype ContT t a = ContT { unContT :: forall r . (a -> t r) -> t r }

instance Monad (ContT t)  where   return x = ContT ($x)   m >>= f  = ContT (\k -> unContT m (\x -> unContT (f x) k))

Both the `mtl` package and the `transformers` package use the same `Monad` instance for their `ContT` type but require `t` to be an instance of `Monad`. Why? [^1]

Maybe because this is needed to prove monad laws?

<snip derivation>

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So, that wasn't the reason. It really is a monad.

In general, ContT r m a is equivalent to Cont (m r) a, and their corresponding Monad instances are also equivalent. But Cont r is a monad for any r, which implies that ContT r m must be a monad for any r and m.

But it was a nice little exercise, right? :-) BTW, is this (ContT t) somehow related to the 'free monad' over t? Cheers Ben