On Mon, May 10, 2010 at 5:51 AM, Milind Patil
For a function
f :: a -> m b f = undefined
I am having trouble understanding how the type of
(>>= f)
is
(>>= f) :: m a -> m b
where, by definition, type of (>>=) is
(>>=) :: (Monad m) => m a -> (a -> m b) -> m b
I do not see how (>>= f) even unifies.
I mean if I code a function with the same type as (>>=) ie.
tt :: (Monad m) => m a -> (a -> m b) -> m b tt = undefined
type of (tt f) does not infer to the same type as (>>= f), from ghc ...
(tt f) :: (Monad ((->) b)) => (m a -> b -> b1) -> b -> b1
There seems to something special about (>>=) apart from its type. And whats (Monad ((->) b))? I am new to Haskell and I may have gaps in my understanding of type inference in Haskell.
It's because >>= is a binary operator. When you partially apply a
binary operator, you get a "section" which applies one of the two
arguments.
Specifically, you have:
(>>=) = \m f -> m >>= f
(m >>=) = \f -> m >>= f
(>>= f) = \m -> m >>= f
There's more in the Haskell tutorial (section 3.2.1)
http://www.haskell.org/tutorial/functions.html
Or you can check the Haskell Report, section 3.5:
http://www.haskell.org/onlinereport/exps.html#sections
--
Dave Menendez