oops, I take that back. It only appears to work if you are sloppy: x :: (Monad m) => m a x = undefined y :: (Monad m) => m b y = undefined g :: (Monad m) => a -> b -> m c g = undefined ex1 :: (Monad m) :: m c ex1 = (g =<< x) =<< y But, if you try to pin down the types you find it only works because they are not all the same Monad m. a :: (Int -> Int) a = return 1 b :: Maybe Int b = Just 2 h :: Int -> Int -> Maybe Int h a b = return (a + b) ex5 :: Maybe Int ex5 = (h =<< a) =<< b :) j. At Thu, 12 Feb 2009 18:04:45 -0600, Jeremy Shaw wrote:
Hello,
You could do:
(f =<< x) =<< y
?
- jeremy
At Thu, 12 Feb 2009 23:36:19 +0000, Edsko de Vries wrote:
Hi,
I can desugar
do x' <- x f x'
as
x >>= \x -> f x'
which is clearly the same as
x >>= f
However, now consider
do x' <- x y' <- y f x' y'
desugared, this is
x >>= \x -> y >>= \y' -> f x' y'
I can simplify the second half to
x >>= \x -> y >>= f x'
but now we are stuck. I feel it should be possible to write something like
x ... y ... f
or perhaps
f ... x ... y
the best I could come up with was
join $ return f `ap` x `ap` y
which is not terrible but quite as easy as I feel this should be. Any hints?
Edsko _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
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