Inventing a suitable infix operator symbol for liftM

Question

When working with monadic expressions in Haskell, the use of liftM's (even in infix position) often seems quite unaesthetic and verbose to me.

Most other monadic primitives (>>=, >>) and even liftM's pure pendant $ are infix operators. This makes me think why there is no operator symbol for monadic lifting.

Do you have reasonable, consistent suggestions for an operator symbol (or why there shouldn't be one)? (I thought of >- and -< (shifting the monad through a function), but they seem to have different meanings in the context of arrows.)

Answer 1

You can use the <$> operator from Control.Applicative.

EDIT: Unfortunately <$> only works for some Monads which are instances of Applicative. Defining some instance Monad m => Applicative m is not possible as this would overlap with the existing Applicative instances IO, Maybe and [].

Answer 2

I'd consider something like <$ (<<$ for liftM2, <<<$ for liftM3).

Answer 3

Since liftM == fmap and (.) == fmap for functions (the ((->) r) instance of Functor), I suggest hiding (.) from Prelude and defining:

import Prelude hiding ((.))
import Control.Monad.Instances () -- import Functor instance for functions

infixr 9 .    
(.) :: (Functor f) => (a -> b) -> (f a -> f b)
(.) = fmap

Examples:

foo = (\x -> x ^ 2) . (\x -> x + 1)
(+1) . [1, 2, 3]
someIO =<< read . getLine -- more useful in do notation because of following
someIO . read =<< getLine -- equivalent to above but uses different instance