On Wed, Jul 14, 2010 at 7:33 AM, prad
xtract p c = do let r = c=~p::[[String]] return r!!0!!0!!1
return is just a function in Haskell, and given that function application has priority over any operator, this mean your last line is :
(return r) !! 0 !! 0 !! 1
return x = [x] so in this case your last line put r in a singleton list : [r] !! 0 !! 0 !! 1
return type is "(Monad m) => a -> m a" where m is a type constructor variable (a type constructor is a parametrized type, think array parametrized on the element type in most language, or template in C++) constrained to be an instance of the Monad typeclass. Monad is an important typeclass in Haskell but here it's enough to look at (!!) type "[a] -> Int -> a" to see that (return r) should be of a list type, list is a monad so this code typecheck, but return is perfectly redundant : For the list monad : then (!! 0) extract r
r !! 0 !! 1 And you get what you wanted in the first place...
So you could write just "r !! 0 !! 1" instead of "return r!!0!!!0!!1" for the same result. In fact xtract could be written :
xtract p c = (c =~ p :: [[String]]) !! 0 !! 1
do-notation (which is just syntactic sugar to write monadic code easily) and return (which is just a function, not a keyword) are only useful when you're working with monads, if you're not you shouldn't use them. -- Jedaï