[Haskell-beginners] Understanding the function monad ((->) r)

[Benjamin Edwards]

The thing that you might also be missing is that function application binds
tightest. Hopefully the parenthesis that Rahul has added help you out
there. If not:

\w -> f (h w) w

f will be applied to the result of (h r) which yields another function,
which is then applied to r

that is

\w ->
let x = h w
g = f x
in g w

would yield exactly the same result. I apologise for the indentation, I
need a better mail client.

Ben

On Tue, 21 Feb 2017 at 14:34 Rahul Muttineni <rahulmutt at 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 at 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 at web.de
> <mailto:50295 at 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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