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

[Olumide]

Thanks a bunch. Simply gorgeous answer.

- Olumide


On 21/02/17 14:32, Rahul Muttineni 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
> <mailto: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>
>         <mailto: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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