Re: [Haskell-cafe] foldr (.) id

sorry for the buggy code

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

On Fri, Oct 26, 2012 at 2:11 PM, Greg Fitzgerald wrote:

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 wrote:

...

...

...

...

...

> Greg Fitzgerald 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