On May 10, 2010, at 05:51 , Milind Patil wrote:
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.
Everyone else having answered the first question, I'll tak
On Mon, May 10, 2010 at 5:51 AM, 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
(>>= 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
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 th
