[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] ?
-}
main=do
-- print result of all implementations to show that they are equal
sequence [ print x| x <- [[1,3], -- [1,3] should be the result
list1,
list1rewritten,
list2,
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.... ;-)
Marc
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