[Haskell-cafe] f^n for functional iteration
doug at cs.dartmouth.edu
Thu Dec 12 03:37:43 UTC 2013
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).
> On Tue, 10 Dec 2013 at 10:51 AM, Danny Gratzer <danny.gratzer at gmail.com> 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 <doug at cs.dartmouth.edu>wrote:
> > 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
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