Exercise 9.8

power :: (a -> a) -> Int -> a -> a
power f 0 = (\x -> x)
power f n = (\x -> f (power f (n-1) x))

Since the function is called power, there is one obvious application that springs to mind, i.e. raising to a power:

raise :: Int -> Int -> Int
raise x y = power (*x) y 1

2 Responses to “Exercise 9.8”

  1. Markus Dittrich says:

    The following is a possible definition for power using
    higher-order functions:

    power :: ( a -> a ) -> Int -> a -> a
    power f n = let funcList = replicate n f
                in (foldr (.) id funcList)
  2. Ricky Liu says:

    It can be simplified to:

    power f n = foldr (.) id (replicate n f)

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