[Haskell-cafe] Repeated function application

[Thomas Hartman]

On second thought... never mind.

The only thing of (somewhat marginal) interest that my latest comment
adds is that the second argument doesn't need to be strict.

Otherwise my code is exactly identical to Dan's.

2008/2/22, Thomas Hartman <tphyahoo at gmail.com>:
> This was easier for me to understand when written so, with the start
>
>  value explicit
>
>   times3 :: (a -> a) -> Int -> (a -> a)
>   times3 f n !start | n == 0 = start
>                        | otherwise = times3 f (n-1) (f start)
>
>   -- no stack overflow :)
>   tTimes3 = times3 (+1) 1000000 0
>
>   Here, only the third arg, the start value, needs to be
>
>  "bangified/strictified", and it's pretty clear why. Without the bang
>  pattern, it stack overflows.
>
>
>   What I'm not sure of is whether this version is in fact completely
>   equivalent to Dan's version above.
>
>   I hope it is.
>
>   2008/2/21, Dan Weston <westondan at imageworks.com>:
>
>  > 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
>   >
>   >
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