5
votes

Suppose that in a Haskell program I have some data whose type is something like:

  • IO [ IO (Int, String, Int) ], or
  • IO [ (Int, String, IO Int) ], or
  • [ (Int, String, IO Int) ]

but I have pure functions that should operate on [ (Int, String, Int) ]. It seems that I'd have to clumsily remove the inside values from the IO monad, until I got something like IO [ (Int, string, Int) ] and then (from inside the IO monad) apply the pure functions. There is no easy pre-defined way to do this, I suppose? Something that would lift a whole data structure into a monad, turning all inside types into pure types? (That would be very convenient!)

2
Thank you guys for the excellent answers! You were absolutely helpful!Jay

2 Answers

6
votes

You could use the liftM* function from the Control.Monad module, or the liftA* functions for applicatives.

liftM allows you to lift a pure function to work inside a Monad, e.g.:

ghci> let s = return "Hello" :: IO String
ghci> liftM reverse s
"olleH"

This way you don't have to manually write things like "s >>= \x -> return (reverse x)" everywhere.

Although, this won't help you with your [(String, Int, IO Int)] example, if the pure function you have deals with a [(String, Int, Int)]. Since the third element in the tuple really isn't an Int.

In that case I'd suggest to first write a function [(String, Int, IO Int)] -> IO [(String, Int, Int)] and that apply the lifted pure function.


This is the most general function I could come up with to do this:

conv :: Monad m => (f (m a) -> m (f a)) -> [f (m a)] -> m [f a]
conv f = sequence . map f

You can call it like so:

liftTrd :: Monad m => (a, b, m c) -> m (a, b, c)
liftTrd (x, y, mz) = mz >>= \z -> return (x, y, z)

conv liftTrd [("hi", 4, return 2)] :: IO [(String, Int, Int)]

This function will only work if you have a single monad that's somewhere deep in a type. If you have multiple, I think you should really be thinking about the type your working in with and see if you can't make it simpler.

4
votes

First some usage example for the solution below called reduce (unless you suggest a better name):

> reduce [(["ab", "c"], "12")] :: [(String, String)]
[("ab","12"),("c","12")]

> reduce [(["ab", "c"], "12")] :: [(Char, Char)]
[('a','1'),('a','2'),('b','1'),('b','2'),('c','1'),('c','2')]

> reduce [("ab", "12"), ("cd", "3")] :: [(Char, Char)]
[('a','1'),('a','2'),('b','1'),('b','2'),('c','3'),('d','3')]

Your example is also solved with it:

complexReduce :: Monad m => m (m (a, b, m [m (c, m d)])) -> m (a, b, [(c, d)])
complexReduce = reduce

And the implementation of reduce:

{-# LANGUAGE FlexibleContexts, FlexibleInstances, IncoherentInstances, MultiParamTypeClasses, UndecidableInstances #-}

import Control.Monad

-- reduce reduces types to simpler types,
-- when the reduction is in one of the following forms:
-- * make a Monad disappear, like join
-- * move a Monad out, like sequence
-- the whole magic of Reduce is all in its instances
class Reduce s d where
  reduce :: s -> d

-- Box is used only for DRY in Reduce instance definitions.
-- Without it we, a Reduce instance would need
-- to be tripled for each variable:
-- Once for a pure value, once for a monadic value,
-- and once for a reducable value
newtype Box a = Box { runBox :: a }
instance Monad m => Reduce (Box a) (m a) where
  reduce = return . runBox
instance Reduce a b => Reduce (Box a) b where
  reduce = reduce . runBox
redBox :: Reduce (Box a) b => a -> b
redBox = reduce . Box

-- we can join
instance (Monad m
  , Reduce (Box a) (m b)
  ) => Reduce (m a) (m b) where
  reduce = join . liftM redBox

-- we can sequence
-- * instance isnt "Reduce [a] (m [b])" so type is always reduced,
--   and thus we avoid overlapping instances.
-- * we cant make it general for any Traversable because then
--   the type system wont find the right patterns.
instance (Monad m
  , Reduce (Box a) (m b)
  ) => Reduce (m [a]) (m [b]) where
  reduce = join . liftM (sequence . fmap redBox)

instance (Monad m
  , Reduce (Box a) (m c)
  , Reduce (Box b) (m d)
  ) => Reduce (a, b) (m (c, d)) where
  reduce (a, b) = liftM2 (,) (redBox a) (redBox b)

instance (Monad m
  , Reduce (Box a) (m d)
  , Reduce (Box b) (m e)
  , Reduce (Box c) (m f)
  ) => Reduce (a, b, c) (m (d, e, f)) where
  reduce (a, b, c) =
    liftM3 (,,) (redBox a) (redBox b) (redBox c)