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Today's Topics:
1. Re: Functions as Applicatives (Olumide)
2. Re: Functions as Applicatives (Tony Morris)
3. Re: Functions as Applicatives (Imants Cekusins)
----------------------------------------------------------------------
Message: 1
Date: Tue, 23 Aug 2016 12:28:09 +0100
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] Functions as Applicatives
Message-ID: <[email protected]>
Content-Type: text/plain; charset=utf-8; format=flowed
I must be missing something. I thought f accepts just one argument.
- Olumide
On 23/08/2016 00:54, Theodore Lief Gannon wrote:
> Yes, (g x) is the second argument to f. Consider the type signature:
>
> (<*>) :: Applicative f => f (a -> b) -> f a -> f b
>
> In this case, the type of f is ((->) r). Specialized to that type:
>
> (<*>) :: (r -> a -> b) -> (r -> a) -> (r -> b)
> f <*> g = \x -> f x (g x)
>
> Breaking down the pieces...
> f :: r -> a -> b
> g :: r -> a
> x :: r
> (g x) :: a
> (f x (g x)) :: b
>
> The example is made a bit confusing by tossing in an fmap. As far as the
> definition above is concerned, 'f' in the example is ((+) <$> (+3)) and
> that has to be resolved before looking at <*>.
>
>
> On Mon, Aug 22, 2016 at 9:07 AM, Olumide <[email protected]
> <mailto:[email protected]>> wrote:
>
> Hi List,
>
> I'm struggling to relate the definition of a function as a function
>
> instance Applicative ((->) r) where
> pure x = (\_ -> x)
> f <*> g = \x -> f x (g x)
>
> with the following expression
>
> ghci> :t (+) <$> (+3) <*> (*100)
> (+) <$> (+3) <*> (*100) :: (Num a) => a -> a
> ghci> (+) <$> (+3) <*> (*100) $ 5
> 508
>
> From chapter 11 of LYH http://goo.gl/7kl2TM .
>
> I understand the explanation in the book: "we're making a function
> that will use + on the results of (+3) and (*100) and return that.
> To demonstrate on a real example, when we did (+) <$> (+3) <*>
> (*100) $ 5, the 5 first got applied to (+3) and (*100), resulting in
> 8 and 500. Then, + gets called with 8 and 500, resulting in 508."
>
> The problem is that I can't relate that explanation with the
> definition of a function as an applicative; especially f <*> g = \x
> -> f x (g x) . Is (g x) the second argument to f?
>
> Regards,
>
> - Olumide
> _______________________________________________
> Beginners mailing list
> [email protected] <mailto:[email protected]>
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
> <http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners>
>
>
>
>
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>
------------------------------
Message: 2
Date: Tue, 23 Aug 2016 21:39:43 +1000
From: Tony Morris <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] Functions as Applicatives
Message-ID: <[email protected]>
Content-Type: text/plain; charset="utf-8"
All functions in Haskell always take one argument.
On 23/08/16 21:28, Olumide wrote:
> I must be missing something. I thought f accepts just one argument.
>
> - Olumide
>
> On 23/08/2016 00:54, Theodore Lief Gannon wrote:
>> Yes, (g x) is the second argument to f. Consider the type signature:
>>
>> (<*>) :: Applicative f => f (a -> b) -> f a -> f b
>>
>> In this case, the type of f is ((->) r). Specialized to that type:
>>
>> (<*>) :: (r -> a -> b) -> (r -> a) -> (r -> b)
>> f <*> g = \x -> f x (g x)
>>
>> Breaking down the pieces...
>> f :: r -> a -> b
>> g :: r -> a
>> x :: r
>> (g x) :: a
>> (f x (g x)) :: b
>>
>> The example is made a bit confusing by tossing in an fmap. As far as the
>> definition above is concerned, 'f' in the example is ((+) <$> (+3)) and
>> that has to be resolved before looking at <*>.
>>
>>
>> On Mon, Aug 22, 2016 at 9:07 AM, Olumide <[email protected]
>> <mailto:[email protected]>> wrote:
>>
>> Hi List,
>>
>> I'm struggling to relate the definition of a function as a function
>>
>> instance Applicative ((->) r) where
>> pure x = (\_ -> x)
>> f <*> g = \x -> f x (g x)
>>
>> with the following expression
>>
>> ghci> :t (+) <$> (+3) <*> (*100)
>> (+) <$> (+3) <*> (*100) :: (Num a) => a -> a
>> ghci> (+) <$> (+3) <*> (*100) $ 5
>> 508
>>
>> From chapter 11 of LYH http://goo.gl/7kl2TM .
>>
>> I understand the explanation in the book: "we're making a function
>> that will use + on the results of (+3) and (*100) and return that.
>> To demonstrate on a real example, when we did (+) <$> (+3) <*>
>> (*100) $ 5, the 5 first got applied to (+3) and (*100), resulting in
>> 8 and 500. Then, + gets called with 8 and 500, resulting in 508."
>>
>> The problem is that I can't relate that explanation with the
>> definition of a function as an applicative; especially f <*> g = \x
>> -> f x (g x) . Is (g x) the second argument to f?
>>
>> Regards,
>>
>> - Olumide
>> _______________________________________________
>> Beginners mailing list
>> [email protected] <mailto:[email protected]>
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>> <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
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Message: 3
Date: Tue, 23 Aug 2016 14:00:01 +0200
From: Imants Cekusins <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Functions as Applicatives
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> I thought f accepts just one argument.
if f is (+) then
f::a -> a -> a
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