[Haskell-beginners] Monad and bind operator interpretation

[Joel Williamson]

This function is actually in the Prelude as (=<<).

On Mon, 14 Dec 2015, 13:17 Dániel Arató <exitconsole at gmail.com> wrote:

> On 14/12/2015, Raja <rajasharan at gmail.com> wrote:
> > So extending this interpretation - can I swap the two parameters (?)
> >
> > Now my new hypothetical interpretation becomes:
> >
> > (>>=) :: (a -> m b) -> m a -> m b
>
> Sure,
> bind' :: Monad m => (a -> m b) -> m a -> m b
> bind' = flip (>>=)
>
> > If i further add parens:
> >
> > (>>=) :: (a -> m b) -> (m a -> m b)
>
> Yeah, that's exactly the same thing. Types are right associative.
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