Hi Dave, I wrote the first one sometime last year and it seemed a suitably simple example for applicative-izing to cement the ideas in some of the code I've been going though in "Learn You a Haskell ..." Interesting stuff. Onward and upward. Thanks, Michael --- On Sat, 9/4/10, David Menendez wrote: From: David Menendez Subject: Re: [Haskell-cafe] On to applicative To: "michael rice" Cc: haskell-cafe@haskell.org Date: Saturday, September 4, 2010, 2:23 PM On Sat, Sep 4, 2010 at 2:06 PM, michael rice wrote: The two myAction functions below seem to be equivalent and, for this small case, show an interesting economy of code, but being far from a Haskell expert, I have to ask, is the first function as small (code wise) as it could be? Michael import Control.Applicative data Color     = Red     | Blue     | Green     | Yellow     | Orange     | Brown     | Black     | White     deriving (Show, Read, Eq, Enum, Ord, Bounded) -- myAction :: IO Color -- myAction = getLine --            >>= \str -> return (read str :: Color) First, you don't need the type annotation here. Haskell will infer it from the annotation on myAction. (You also don't need the type on myAction, but that's not important now.) myAction = getLine >>= \str -> return (read str) Second, you have the pattern \x -> f (g x), so you can replace the lambda with function composition. myAction = getLine >>= return . read Third, there is a standard function liftM, defined in Control.Monad, where liftM f m = m >>= return . f, so myAction = liftM read getLine Finally, we expect an instance of Monad to also be an instance of Functor, with fmap = liftM, and we also have f <$> m = fmap f m, so  myAction :: IO Color myAction = read <$> getLine -- Dave Menendez http://www.eyrie.org/~zednenem/