(>>= f) is equivalent to (flip (>>=) f), not to ((>>=) f). You can try this with your own function this way: (&$^) :: (Monad m) => m a -> (a -> m b) -> m b (&$^) = undefined :t (&$^ f) Milind Patil wrote:
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.
regards, Milind Patil
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