Re: [Haskell-beginners] Monad and bind operator interpretation

This function is actually in the Prelude as (=<<). On Mon, 14 Dec 2015, 13:17 Dániel Arató wrote:

On 14/12/2015, Raja 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. _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners