Actually, the second argument is already strict, and the ! doesn't make
it any stricter (and is therefore gratuitous): when you evaluate the
conditional (n == 0), n is evaluated.
Dan
Thomas Hartman wrote:
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 <[EMAIL PROTECTED]>:
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 <[EMAIL PROTECTED]>:
> 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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