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Today's Topics:
1. Re: Understanding the function monad ((->) r) (Rahul Muttineni)
2. Re: Understanding the function monad ((->) r) (Benjamin Edwards)
3. Re: Understanding the function monad ((->) r) (Olumide)
----------------------------------------------------------------------
Message: 1
Date: Tue, 21 Feb 2017 20:02:55 +0530
From: Rahul Muttineni <[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:
<CANij+eTi1ZKyWytgu84VPJh0P1=b_+zhi-6r-m1r22jv-ec...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
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]> 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]>> 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
>>
>> _______________________________________________
>> Beginners mailing list
>> [email protected] <mailto:[email protected]>
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
>>
>>
>> _______________________________________________
>> Beginners mailing list
>> [email protected]
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
>>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
--
Rahul Muttineni
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Message: 2
Date: Tue, 21 Feb 2017 15:08:45 +0000
From: Benjamin Edwards <[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:
<CAN6k4njnviJo6fPcx33WQdEN=ibyq3fvehykwns81ytw+9h...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
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 <[email protected]> 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]> 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]>> 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
>
> _______________________________________________
> Beginners mailing list
> [email protected] <mailto:[email protected]>
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
>
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
>
>
>
> --
> Rahul Muttineni
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
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Message: 3
Date: Tue, 21 Feb 2017 15:52:22 +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
On 21/02/2017 15:08, Benjamin Edwards wrote:
> 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
Did you mean to write (h w)?
- Olumide
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