14 Jul
2013
14 Jul
'13
1:36 a.m.
Hi, Brent, thanks for pointing it out, it helps.
Regards!
-
wudeng
On Sun, Jul 14, 2013 at 6:58 AM, Brent Yorgey
On Sat, Jul 13, 2013 at 05:47:45PM +0800, Deng Wu wrote:
It makes sense to me, so I change the definition of cps to something like this:
cfold f z [] = z cfold f z (x:xs) = (\y -> cfold f y xs) (f x z)
Note that if we just reduce the application of (\y -> ...) to (f x z), we get
cfold f z (x:xs) = cfold f (f x z) xs
But that is just the usual definition of foldl. So it certainly works fine but it is not in CPS.
-Brent
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