monads

I see everywhere that Applicative can handle side effects, but all the simple examples I've seen are just combining stuff together like: > (,,) <$> [1,2] <*> ["a", "b", "c"] <*> ["foo", "bar"] [(1,"a","foo"),(1,"a","bar"),(1,"b","foo"),(1,"b","bar"), (1,"c","foo"),(1,"c","bar"),(2,"a","foo"),(2,"a","bar"), (2,"b","foo"),(2,"b","bar"),(2,"c","foo"),(2,"c","bar")] Which is cool but I can't see how that links to side effects. My understanding is that Applicative is a weak monad and so you can handle side effects (as you would do with a State monad) but you can't reuse the result of the previous

I'm having some real trouble designing the counterfunction of Haskell's sequence function, which Hoogle tells me doesn't yet exist. This is how it behaves: ghci> sequence [Just 7, Just 8, Just 9] Just [7,8,9] ghci> sequence [getLine, getLine, getLine] hey there stack exchange ["hey","there","stack exchange"] :: IO [String] My problem is making a function like this: unsequence :: (Monad m) => m [a] -> [m a] So that it behaves like this: ghci> unsequence (Just [7, 8, 9]) [Just 7, Just 8, Just 9] ghci> sequence getLine hey ['h','e','y'] :: [IO Char] --(This would actually cause an error, but hey

I'm currently porting some code from traditional Scala to Scalaz style. It's fairly common through most of my code to use the Seq trait in my exposed API signatures rather than a concrete type (i.e. List, Vector) directly. However, this poses some problem with Scalaz, since it doesn't provide an implementation of a Bind[Seq] typeclass. i.e. This will work correctly. List(1,2,3,4) >>= bindOperation But this will not Seq(1,2,3,4) >>= bindOperation failing with the error could not find implicit value for parameter F0: scalaz.Bind[Seq] I assume this is an intentional design decision in Scalaz -

I've frequently heard claims that Haskell doesn't have variables; in particular, this answer claims that it doesn't, and it was upvoted at least nine times and accepted. So does it have variables or not, and why? This question also appears to apply ML, F#, OCaml, Erlang, Oz, Lava, and all SSA intermediate languages. Don Stewart Haskell has immutable variables (variables in the math sense) by default: foo x y = x + y * 2 By default variables are not mutable cells . Haskell also has mutable cells though, but you enable them explicitly: > import Data.IORef (newIORef, readIORef, writeIORef) > v <-

So far, every monad (that can be represented as a data type) that I have encountered had a corresponding monad transformer, or could have one. Is there such a monad that can't have one? Or do all monads have a corresponding transformer? By a transformer t corresponding to monad m I mean that t Identity is isomorphic to m . And of course that it satisfies the monad transformer laws and that t n is a monad for any monad n . I'd like to see either a proof (ideally a constructive one) that every monad has one, or an example of a particular monad that doesn't have one (with a proof). I'm interested