Exercise 9.5

applyAll :: [(a -> a)] -> a -> a
applyAll [] x = x
applyAll (f:fs) x = f (applyAll fs x)

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4 Responses to “Exercise 9.5”

  1. anonyoumous says:
    21 February 2008 at 12:48 pm

    applyAll fs x = foldr id x fs

  2. anonymous says:
    10 June 2008 at 11:15 pm

    applyAll fs x = foldr ($) x fs

  3. Kazu Yamamoto says:
    6 February 2009 at 12:19 am

    applyAll = foldr (.) id

  4. RJP says:
    22 June 2014 at 8:45 am

    id, ($), (.) and \ aren’t introduced until later in the text, using concepts only from 9.1 and earlier, one answer would be:

    applyAll :: [(a -> a)] -> a -> a
    applyAll = flip (foldr aux)
    where aux f x = f x

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