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Today's Topics:
1. Re: Where is the accumulator in the expression foldr (<=<)
return (replicate x oveKnight)? -- from LYAH example (Tony Morris)
----------------------------------------------------------------------
Message: 1
Date: Fri, 27 Jul 2018 13:29:31 +1000
From: Tony Morris <tonymor...@gmail.com>
To: beginners@haskell.org
Subject: Re: [Haskell-beginners] Where is the accumulator in the
expression foldr (<=<) return (replicate x oveKnight)? -- from LYAH
example
Message-ID: <e40e24f6-90a7-edc4-c267-54d2a64e8...@gmail.com>
Content-Type: text/plain; charset="utf-8"
foldr does not involve an accumulator. Only left folds involve an
accumulator. Specifically, the expression:
foldl f z list
can be though of as this loop:
var r = z
foreach(element in list) {
r = f(r, element)
}
return r
That is why, for example, foldl (\r element -> element : r) [] will
reverse a list. The accumulator here is the (r) value in the loop.
The foldr function does constructor replacement, and in no prescribed
order. The expression foldr f z list will replace every (:) with (f) and
[] with (z) in list.
So the expression foldr (<=<) return will take a list, such as thing one:
let list = a : b : c : d : e : []
and turn it into this value:
foldr (<=<) return list
=
a <=< b <=< c <=< d <=< e <=< return
This will occur in no prescribed order, although the replacement is
right-associative. It is often said that foldr "starts at the
right-most", however, this is untrue, since foldr works on infinite
list, which has no notion of right-most. Importantly, there is no notion
of "accumulator" here, only constructor replacement.
The right-associativity simply means that the parentheses are to the right:
a <=< (b <=< (c <=< (d <=< (e <=< return))))
However, this does not necessarily impose an execution order, or
accumulator. Consider this expression:
foldr const (repeat 1)
This will produce the value:
1 `const` (1 `const` (1 `const` 1 ...
The result of normalising this expression produces the value 1. There is
no notion of the "right-most" to "accumulate" anything. It simply
evaluates and the answer is 1.
On 07/26/2018 11:44 AM, Olumide wrote:
> 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
> Beginners@haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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