Converting functions to CPS

Question

I have the following CPS implementation of until:

until' p f x cc = p x (\y -> if y then cc(x)
                             else until' p f (f x (\z -> cc(z))) cc)

Which type checks ok! Now, trying to CPS map:

map' f (x:xs) cc = f x (\y -> map f xs (\ys -> cc(y:ys)))

Another possible implementation:

map' f (x:xs) cc = cc(f x (\y cc' -> map f xs (\ys -> cc'(y:ys))))

However none of them type checks. Where am I doing the mistake?

    Couldn't match expected type ‘([a1] -> t2) -> t1’
                with actual type ‘[(a1 -> t1) -> t]’
    Relevant bindings include
      y :: a1 (bound at test.hs:6:26)
      cc :: [a1] -> t2 (bound at test.hs:6:15)
      f :: a -> (a1 -> t1) -> t (bound at test.hs:6:6)
      map' :: (a -> (a1 -> t1) -> t) -> [a] -> ([a1] -> t2) -> t
        (bound at test.hs:6:1)
    The function ‘map’ is applied to three arguments,
    but its type ‘(a -> (a1 -> t1) -> t) -> [a] -> [(a1 -> t1) -> t]’
    has only two
    In the expression: map f xs (\ ys -> cc (y : ys))
    In the second argument of ‘f’, namely
      ‘(\ y -> map f xs (\ ys -> cc (y : ys)))’
Failed, modules loaded: none.

You are calling `map` instead of `map'` in the recursive call. — chi, Oct 24 '15 at 21:23
You shouldn't edit the answer into your question - I got all confused at first because your first version now works... — Ørjan Johansen, Oct 25 '15 at 05:44

score 4 · Answer 1 · answered Oct 24 '15 at 19:50

4

When rewriting in CPS the continuation takes the result, so if you write the expected type of your map function you have:

mapk :: (a -> b) -> [a] -> ([b] -> r) -> r

so the continuation takes the result list as a parameter. If you look at your implementation:

map' f (x:xs) cc = f x (\y -> map f xs (\ys -> cc(y:ys)))

y and ys should have the same type ([b]) however you are trying to combine them with (:) which expects a b and [b]. Instead you want something like:

mapk _ [] k = k []
mapk f (x:xs) k = mapk f xs (\rs -> k $ (f x):rs)

answered Oct 24 '15 at 19:50

Lee

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`f` presumably should be CPS converted as well, for the complete transform. – dfeuer Oct 24 '15 at 20:12
@dfeuer could you give a implementation that CPS `f` ? – cybertextron Oct 24 '15 at 20:42

score 0 · Answer 2 · answered Oct 24 '15 at 21:41

0

It was a typo on

map' f (x:xs) cc = f x (\y -> map f xs (\ys -> cc(y:ys)))

should be:

map' f (x:xs) cc = f x (\y -> map' f xs (\ys -> cc(y:ys)))

answered Oct 24 '15 at 21:41

cybertextron

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Converting functions to CPS

2 Answers2