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Today's Topics:
1. Re: Understanding the function monad ((->) r) (Olumide)
----------------------------------------------------------------------
Message: 1
Date: Sat, 25 Feb 2017 20:19:15 +0000
From: Olumide <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Understanding the function monad
((->) r)
Message-ID: <[email protected]>
Content-Type: text/plain; charset=utf-8; format=flowed
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 <[email protected]
> <mailto:[email protected]>> 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 <[email protected]
> <mailto:[email protected]>
> <mailto:[email protected] <mailto:[email protected]>>> 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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