[Haskell-cafe] f^n for functional iteration
me at lelf.lu
Fri Dec 13 14:59:26 UTC 2013
Doug McIlroy <doug at cs.dartmouth.edu> writes:
> 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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