Ben Butler-Cole writes:

times :: (a -> a) -> Int -> (a -> a) times f 0 = id times f n = f . (times f (n-1))

Am I missing something more general ...I can't help feeling that there must be a way to get rid of the explicit recursion.

How would you implement times?

Anything against (apart an auxiliary list, and "x" not curried away) times n f x = (iterate f x)!!n Jerzy Karczmarczuk

Ben Butler-Cole wrote:

Hello

I was surprised to be unable to find anything like this in the standard libraries:

times :: (a -> a) -> Int -> (a -> a) times f 0 = id times f n = f . (times f (n-1))

Am I missing something more general which would allow me to repeatedly apply a function to an input? Or is this not useful?

Invariably, this seems to invite a stack overflow when I try this (and is usually much slower anyway). Unless f is conditionally lazy, f^n and f will have the same strictness, so there is no point in keeping nested thunks. If you apply f immediately to x, there is no stack explosion and faster runtime: times :: (a -> a) -> Int -> (a -> a) times f !n !x | n > 0 = times f (n-1) (f x) | otherwise = x Dan