[Haskell-cafe] foldr (.) id

[Greg Fitzgerald]

Hmm, neato.  but didn't make life any easier!

Data.Monoid> (appEndo . mconcat . map Endo) [(+10), (+20)] 3
33
Data.Monoid> (foldr (.) id) [(+10), (+20)] 3
33

I had hoped for something like:

> mconcat [(+10), (+20)] 3

But I suppose that's nonsense, considering this works:

> mconcat [(++"10"), (++"20")] "3"
"310320"


I think this is the most general solution?

import Control.Category
import Data.Foldable
import Prelude hiding (foldr, (.), id)

compose :: (Foldable t, Category cat) => t (cat a a) -> cat a a
compose = foldr (.) id


Usage:

> compose [(+10), (+20)] 3

Real-world use case:

> let parseOrIgnore p = either (const s) id . parse p s
> parseAllOrIgnore = compose . map parseOrIgnore [p1, p2, p3]

Naming:

"(.)/compose" is consistent with "(+)/sum", "(*)/product", "(&&)/and", etc.

Thoughts?

-Greg


On Fri, Oct 26, 2012 at 12:31 PM, John Wiegley <jwiegley at gmail.com> wrote:

> >>>>> Greg Fitzgerald <garious at gmail.com> writes:
>
> > I've recently found myself using the expression: "foldr (.) id" to
> compose a
> > list (or Foldable) of functions.
>
> You want the Endo monoid:
>
>     ghci> appEndo (Endo (+ 10) <> Endo (+ 20)) $ 3
>       33
>
> John
>
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