On 2008 Sep 24, at 22:51, Daryoush Mehrtash wrote:
I am having hard time making sense of the types in the following
example from the Applicative Programming paper: http://www.cs.nott.ac.uk/~ctm/IdiomLite.pdf
ap :: Monad m ⇒ m (a → b ) → m a → m b
ap mf mx = do
f ← mf
x ← mx
return (f x )
Using this function we could rewrite sequence as:
sequence :: [ IO a ] → IO [ a ]
sequence [ ] = return [ ]
sequence (c : cs ) = return (:) 'ap' c 'ap' sequence cs
I am specifically confused over the type of "m" in:
return (:) 'ap' c
"c" is obviously an instance of IO a monad. "return (:)" on the
other hand (at least as I would expect it) is an instance of " ->"
monad.
Note that he first argument to ap is a function wrapped by a monad.
"return (:)" wraps the function/operator (:) in an arbitrary monad
(but then the type signature of sequence makes the monad IO).
(:) :: a -> [a] -> [a]
return (:) :: IO (a -> ([a] -> [a])) -- b is ([a] -> [a])
ap (return (:)) :: IO a -> IO ([a] -> [a]) -- (IO a) is (m a) and
(IO ([a] -> [a])) is (m b)
return (:) `ap` c :: IO ([a] -> [a]) -- c is (IO a) per type
signature
return (:) `ap` c `ap` :: IO [a] -> IO [a] -- f in ap is the
preceding IO ([a] -> [a])
return (:) `ap` c `ap` sequence cs :: IO [a] -- (sequence cs) is
(IO [a]) per type signature
--
brandon s. allbery [solaris,freebsd,perl,pugs,haskell] [EMAIL PROTECTED]
system administrator [openafs,heimdal,too many hats] [EMAIL PROTECTED]
electrical and computer engineering, carnegie mellon university KF8NH
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