Thanks for this response. the Monad instance for ((->) r) has been bugging me as well.

On Tue, Feb 21, 2017 at 6:32 AM, Rahul Muttineni <rahulmutt@gmail.com> wrote:
Hi Olumide,

Let the types help you out.

The Monad typeclass (omitting the superclass constraints):

class Monad m where
  return :: a -> m a
  (>>=) :: m a -> (a -> m b) -> m b

Write out the specialised type signatures for (->) r:

{-# LANGUAGE InstanceSigs #-}
-- This extension allows you to specify the type signatures in instance declarations

instance Monad ((->) r) where
  return :: a -> (r -> a)
  (>>=) :: (r -> a) -> (a -> (r -> b)) -> (r -> b)

Now we look at how to make some definition of return that type checks. We're given an a and we want to return a function that takes an r and returns an a. Well the only way you can really do this is ignoring the r and returning the value you were given in all cases! Because 'a' can be *anything*, you really don't have much else you can do! Hence:
  
  return :: a -> (r -> a)
  return a = \_ -> a

Now let's take a look at (>>=). Since this is a bit complicated, let's work backwards from the result type. We want a function that gives us a b given an r and given two functions with types (r -> a) and (a -> (r -> b)). To get a b, we need to use the second function. To use the second function, we must have an a, which we can get from the first function!

  (>>=) :: (r -> a) -> (a -> (r -> b)) -> (r -> b)
  (>>=) f g = \r -> (g (f r)) r

Hope that helps!
Rahul
  

On Tue, Feb 21, 2017 at 5:04 PM, Olumide <50295@web.de> wrote:
On 21/02/2017 10:25, Benjamin Edwards wrote:
What is it that you are having difficulty with? Is it "why" this is a
good definition? Is it that you don't understand how it works?

I simply can't grok f (h w) w.

- Olumide

On Tue, 21 Feb 2017 at 10:15 Olumide <50295@web.de
<mailto:50295@web.de>> wrote:

    Hello List,

    I am having enormous difficulty understanding the definition of the bind
    operator of ((->) r) as show below and would appreciate help i  this
    regard.

    instance Monad ((->) r) where
         return x = \_ -> x
         h >>= f = \w -> f (h w) w

    Thanks,

    - Olumide

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--
Rahul Muttineni

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