> with the cpsfold I get > > f x1 a (\y1 -> f x2 y1 (\y2 -> f x3 y3 (\y3 -> f x4 y4 (\y4 -> y4) > > so x1 and a are available immediately for f to use, and f x1 a is the > outermost expression so will be evaluated first.
Yes, however, if f just calls its continuation without forcing the evaluation of at least its second argument, e.g. f x y k = k (g x y) you get f x1 a (\y1 -> f x2 y1 (\y2 -> f x3 y3 (\y3 -> f x4 y4 (\y4 -> y4)))) => f x2 (g x1 a) (\y2 -> f x3 y3 (\y3 -> f x4 y4 (\y4 -> y4))) => f x3 (g x2 (g x1 a)) (\y3 -> f x4 y4 (\y4 -> y4))) => f x4 (g x3 (g x2 (g x1 a))) (\y4 -> y4) => g x4 (g x3 (g x2 (g x1 a))) like the foldr. > > larger x y = if x > y then x else y > > cpslarger x y k = if x > y then k x else k y Yes, with this definition of `cpslarger' no stack is used, because the comparison forces evaluation. With cpslarger x y k = k (larger x y) it does not work. Still, if your definition of cpslarger is natural for your application, it is a nice solution of the problem. Ciao, Olaf -- OLAF CHITIL, Dept. of Computer Science, The University of York, York YO10 5DD, UK. URL: http://www.cs.york.ac.uk/~olaf/ Tel: +44 1904 434756; Fax: +44 1904 432767 _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
