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Today's Topics:
1. Where is the accumulator in the expression foldr (<=<)
return (replicate x oveKnight)? -- from LYAH example (Olumide)
2. Re: Where is the accumulator in the expression foldr (<=<)
return (replicate x oveKnight)? -- from LYAH example (Ut Primum)
3. Re: Where is the accumulator in the expression foldr (<=<)
return (replicate x oveKnight)? -- from LYAH example (Paul)
----------------------------------------------------------------------
Message: 1
Date: Thu, 26 Jul 2018 02:44:55 +0100
From: Olumide <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: [Haskell-beginners] Where is the accumulator in the
expression foldr (<=<) return (replicate x oveKnight)? -- from LYAH
example
Message-ID: <[email protected]>
Content-Type: text/plain; charset=utf-8; format=flowed
Dear List,
Chapter 13 of LYAH
(http://learnyouahaskell.com/for-a-few-monads-more#useful-monadic-functions)
has the following code block
import Data.List
inMany :: Int -> KnightPos -> [KnightPos]
inMany x start = return start >>= foldr (<=<) return (replicate x oveKnight)
What I'd like to know is where the accumulator of foldr is in this example.
Regards,
- Olumide
------------------------------
Message: 2
Date: Thu, 26 Jul 2018 08:11:28 +0200
From: Ut Primum <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Where is the accumulator in the
expression foldr (<=<) return (replicate x oveKnight)? -- from LYAH
example
Message-ID:
<canjdmkl+shdsjb6lnufyfki6s06rkbvgcb6b0_ot0wkaer9...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Hi,
Looking at the structure of the expression, you should have
foldr operator accumulator list
So in your example the accumulator should be "return" (because it is the
second argument)
Il gio 26 lug 2018, 03:45 Olumide <[email protected]> ha scritto:
> Dear List,
>
> Chapter 13 of LYAH
> (
> http://learnyouahaskell.com/for-a-few-monads-more#useful-monadic-functions)
>
> has the following code block
>
> import Data.List
>
> inMany :: Int -> KnightPos -> [KnightPos]
> inMany x start = return start >>= foldr (<=<) return (replicate x
> oveKnight)
>
> What I'd like to know is where the accumulator of foldr is in this example.
>
> Regards,
>
> - Olumide
>
>
>
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
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Message: 3
Date: Thu, 26 Jul 2018 09:14:07 +0300
From: Paul <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] Where is the accumulator in the
expression foldr (<=<) return (replicate x oveKnight)? -- from LYAH
example
Message-ID: <[email protected]>
Content-Type: text/plain; charset="utf-8"; Format="flowed"
Hello,
`return` is the initial value. So, `replicate x oveKnignt)` is a list of
functions. `foldr` folds them with initial value `return` with the `<=<`
between them: functions are folding with (<=<). First folding value is
`return` function. You can check the types with :t something in the GHCi.
Result of folding is flow of functions or long functions circuit.
`return start` is the same as to pass `start` to this functions circuit:
inMany x start = foldr (<=<) return (replicate x oveKnight) $ start
Idea seems, to make from [KnighPos -> [KnighPos]] functions list one
function: KnighPos -> [KnighPos] performing those functions step by step
(<=<) and to pass `start` to it. Due to `<=<` joining of functions is
not "do", but "do for each...", because <=< is in the list monad.
Something like:
for x in f-last start:
for y in f-prelast x:
...
You can look at these types:
:t foldr
foldr :: Foldable t => (a -> b -> b) -> b -> t a -> b
:t (<=<)
(<=<) :: Monad m => (b -> m c) -> (a -> m b) -> a -> m c
26.07.2018 04:44, Olumide wrotes:
> Dear List,
>
> Chapter 13 of LYAH
> (http://learnyouahaskell.com/for-a-few-monads-more#useful-monadic-functions)
> has the following code block
>
> import Data.List
>
> inMany :: Int -> KnightPos -> [KnightPos]
> inMany x start = return start >>= foldr (<=<) return (replicate x
> oveKnight)
>
> What I'd like to know is where the accumulator of foldr is in this
> example.
>
> Regards,
>
> - Olumide
>
>
>
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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