You're right! I neglected that there is a apostrophe, I should have been more careful. Thanks for your help! Regards! - wudeng On Sat, Jul 13, 2013 at 9:46 PM, Kim-Ee Yeoh wrote:

On Sat, Jul 13, 2013 at 4:47 PM, Deng Wu wrote:

...

cfold’ f z (x:xs) = f x z (\y -> cfold’ f y xs)

and gives the test result:

CPS> cfold (+) 0 [1,2,3,4]

The apostrophe matters!

Functions cfold and cfold' (read: see-fold prime) are two different functions, for which you gave the definition of one but not the other.

To clear up your confusion, try giving the type signatures for both. Daume's text is normally pretty good about this but the section on cfold falls down in this regard.

Let me do the cfold one for you:

cfold :: (a -> r -> r) -> r -> [a] -> r

-- Kim-Ee

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Hi, Brent, thanks for pointing it out, it helps. Regards! - wudeng On Sun, Jul 14, 2013 at 6:58 AM, Brent Yorgey wrote:

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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