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 _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
Hello Marc,
Sunday, January 08, 2006, 3:19:56 AM, you wrote: MW> list2=do { x <- [1,2,3]; guard (odd x); return x} -- <- provided by xerox list3 = [ x | x <- [1,2,3], odd x] -- Best regards, Bulat mailto:[hidden email] _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
On Sun, Jan 08, 2006 at 12:19:40PM +0300, Bulat Ziganshin wrote:
> Hello Marc, > Sunday, January 08, 2006, 3:19:56 AM, you wrote: > MW> list2=do { x <- [1,2,3]; guard (odd x); return x} -- <- provided by xerox > list3 = [ x | x <- [1,2,3], odd x] list4= take 2 $ [2*x+1,x<-[0..]] ;-) Hi Bulat.. Thanks for your reply. Of cause I know this version, too. My goal has been to understand why the other versions list1,2rewritten give the same result.. (especially which monadic implementation of >>= is used in each case..) I'm trying to understand this simple example because I want to understand some other monad examples beeing more complicated.. See my comments of the last post for details.. If something is unclear.. Please tell me.. Some hints like: look at the defintion of Monad xy should be sufficiant.. Sincerly Marc _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
In reply to this post by Marc Weber
Am Sonntag, 8. Januar 2006 01:19 schrieb Marc Weber:
> 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 ? Let's figure that out (I use Int, although Integral a => a is the most general type). Overall, we have 'ok :: Int -> [b]', from the expression '[1,2,3] >>= ok'. Now in line 1r3 we see that ok x (since that pattern is irrefutable, there's no need for line 1r4) is 'return (odd x) >>= ok2', so by that line alone we can infer the type 'Monad m => Bool -> m c' for ok2. But the result of ok2 is the result of ok, so we find '[b] === m c', hence in line 1r3, >>= is used at type [Bool] -> (Bool -> [b]) -> [b]. Finally, by 1r1 we see that 'b' is the input-type of ok, i.e. Int, so our type-inference has led to list1annotated = let ok :: Int -> [Int] ok x = let ok2 :: Bool -> [Int] ok2 True = return x -- might as well write [x] ok2 _ = fail "ok2" in ((>>=) :: [Bool] -> (Bool -> [Int]) -> [Int]) (return (odd x) :: [Bool]) ok2 in ((>>=) :: [Int] -> (Int -> [Int]) -> [Int]) [1,2,3] ok > > 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? -} > For [] we have list >>= func = concatMap func list, for Maybe it's Just x >>= func = func x Nothing >>= func = Nothing and look at the code for more, I can recommend -- besides the Control.Monad.Whichevers -- ReadP and Parsec. If you've spent some time grasping that, you'll become more comfortable with monads. > > > 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? Exactly. > 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] ? -} mzero is not a constructor (as witnessed by the lowercase spelling), but a special value that 'm a' must contain for a MonadPlus m (and arbitrary a). One of the laws requested for instances of MonadPlus is 'mzero >>= f === mzero'. For lists, this is fulfilled, since concat (map f []) = concat [] = []. If we evaluate the above ok, we have ok 1 = guard (odd 1) >> return 1 = guard True >>= (\_ -> return 1) = return () >>= (\_ -> return 1) = [()] >>= (\_ -> return 1) = concat $ map (\_ -> return 1) [()] = concat $ [return 1] = concat [[1]] = [1] ok 2 = guard (odd 2) >> return 2 = guard False >> return 2 = mzero >>= (\_ -> return 2) = concat $ map (\_ -> return 2) [] = concat [] = [] > > > 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 Hope that helps, Daniel _______________________________________________ Haskell-Cafe mailing list [hidden email] http://www.haskell.org/mailman/listinfo/haskell-cafe |
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