[Haskell-cafe] do {x<-[1,2,3]; True <- return (odd x); return x}.. why? (do notation, monads, guards)

Marc Weber marco-oweber at gmx.de
Sat Jan 7 19:19:56 EST 2006

Here is a simple program implementing the above function in 4 different
ways.. See my comments to get to know where I have problems:

----------  begin test.hs ----------

module Main where

import IO
import Control.Monad.List

{- list1,2 are both implementations of the same function f=[1,3] ;-)
   I've both rewritten with the translation rules for do notation to 
   better understand what's going on and where the differences are

list1=do { x <- [1,2,3]; True <- return (odd x); return x}
list2=do { x <- [1,2,3]; guard (odd x); return x} -- <- provided by xerox

list1rewritten :: [Int]
list1rewritten=let ok x = let ok2 True = do return x  --1r1
			      ok2 _ = fail "ok2"      --1r2
			  in return (odd x) >>= ok2   --1r3
		   ok _ = fail "outer"                --1r4
       in [1,2,3] >>= ok
{- The outer let .. in >>= is used to "call" the inner >>= 
   for each element of [1,2,3] (the list Monad causes this)

   True <- return (odd x): really nice trick...!
   if x is odd then line --1r1 is matched the values is returned
   otherwise        line --1r2 is matched calling fail
				    which is implemented as
                                    = [] ignoring the message hence no element is added
   but I'm not sure which implementation of >>= is used in --lr3:
   It should satisfy (Monad m) => m Bool -> (Bool -> m Int), right ?

   Looking at the definition taken from GHC/Base.lhs:

  class  Monad m  where
      (>>=)       :: forall a b. m a -> (a -> m b) -> m b

  and a sample implementation:
  instance  Monad []  where
      m >>= k             = foldr ((++) . k) [] m

  I wonder how a, b (from m a and m b)  and m (from class Monad m) are renated?
  Can you tell me how the implementation declaration of m a -> (...) -> m b differs in these cases:
  eg: 1. a = Int, b=String 2. the other way round: a=String b=Int?

list2rewritten :: [Int]
list2rewritten = let ok x = guard (odd x) >>  return x
		     ok _ = fail "I think never used?"
		 in [1,2,3] >>= ok 
{- Here ok is feeded with 1,2 and three due to the list Monad again? 
    So fail will never be called, right?
    I also know that guard returns either the monad data type constructor mzero or return ()
    But how is this used in combintation with >> return x::Int to return either [] or [x] ?

  -- print result of all implementations to show that they are equal
  sequence [ print x| x <- [[1,3], -- [1,3] should be the result
			    list2rewritten ] ]

--------------  end -------------------------

I hope there will be some time when I can say: Monads.. I don't bother anymore I'm practicing every night while dreaming.... ;-)


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