Talk:Monad

From HaskellWiki
Jump to navigation Jump to search
The printable version is no longer supported and may have rendering errors. Please update your browser bookmarks and please use the default browser print function instead.

About some of that "Functor/Applicative/Monad" business, a little equivalence formula that's worked well for me goes like so:

    ap (return f) x == (return . f) =<< x == liftM f x == fmap f x

Just to refresh, (=<<) is the reversed bind. Also, 'ap' is the Monad version of the applicative operator, (<*>) - see the Control.Applicative module for more help.

Usually, I stay away from using liftM, since it's identical to fmap, only confined to Monads, as 'ap' is to (<*>). However, the other liftMn functions are a nice generalization of the zipWithn family to Monads. The benefits of that, I'll leave for another discussion.

BMeph 04:30, 11 March 2008 (UTC)

Hope my edit is OK, about Monads as composable computation descriptions whose essence is separation of composition timeline from the run time. That's my hard won insight anyhow. WillNess 12:22, 9 June 2010 (UTC)