On Mon, May 10, 2010 at 5:51 AM, Milind Patil <milind_patil@xxxxxxxxxxx> wrote:
> For a function
> f :: a -> m b
> f = undefined
> I am having trouble understanding how the type of
> (>>= f)
> (>>= 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
> 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
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)
Or you can check the Haskell Report, section 3.5:
Dave Menendez <dave@xxxxxxxxxxxx>
Haskell-Cafe mailing list