applyAll :: [(a -> a)] -> a -> a
applyAll [] x = x
applyAll (f:fs) x = f (applyAll fs x) |
applyAll :: [(a -> a)] -> a -> a
applyAll [] x = x
applyAll (f:fs) x = f (applyAll fs x)
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applyAll fs x = foldr id x fs
applyAll fs x = foldr ($) x fs
applyAll = foldr (.) id
id, ($), (.) 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