Re: [Haskell-cafe] f^n for functional iteration

Doug McIlroy writes: http://hackage.haskell.org/package/NumInstances

Agreeing with the analysis, I will sharpen my question. Is option 2 possible at all, regardless of sanity concerns (e.g. incomplete implementation of Num).

Doug

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On Tue, 10 Dec 2013 at 10:51 AM, Danny Gratzer wrote

Well (^) is already used for their traditional meaning and using this exact operator would require

1. Shadowing (^) from prelude 2. Making (a -> a) an instance of Num (impossible to do sanely)

You can just use a different operator

f .^. n = foldl (.) id $ replicate n f

On Tue, Dec 10, 2013 at 10:45 AM, Doug McIlroy wrote:

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Is there a trick whereby the customary notation f^n for iterated functional composition ((\n f -> foldl (.) id (replicate n f)) n f) can be defined in Haskell?

Doug McIlroy