On May 20, 2009 05:50:54 Claus Reinke wrote:
I'm not at all sure what you're aiming for (the above doesn't compile),
Yeah. The newtype Step a b was required to break the recursive types, and I dropped it when performing the various transformations, so they don't type check. Here it is again with the newtypes so each bit can be compiled. 0- initial code newtype Step a b = Step { unStep :: (a -> Step a b -> b) -> b -> b } iter :: [a] -> (a -> Step a b -> b) -> b -> b iter [] next done = done iter (x:xs) next done = next x (Step $ iter xs) count :: Int -> Char -> Step Char Int -> Int count i x step = unStep step (count $ i+1) (i+1) test :: String -> Int test xs = iter xs (count 0) 0 1a- avoid forming the (count _) closures by passing the function and the argument instead of the function bound to the argument newtype Step a c b = Step { unStep :: (c -> a -> Step a c b -> b) -> c -> b -> b } iter :: [a] -> (c -> a -> Step a c b -> b) -> c -> b -> b iter [] next i done = done iter (x:xs) next i done = next i x (Step $ iter xs) count :: Int -> Char -> Step Char Int Int -> Int count i x step = unStep step count (i+1) (i+1) test :: String -> Int test xs = iter xs count 0 0 1b- avoid forming the (iter _) closure by passing the function and the argument instead of the function bound to the argument newtype Step a c b = Step { unStep :: [a] -> (c -> a -> Step a c b -> [a] -> b) -> c -> b -> b } iter :: [a] -> (c -> a -> Step a c b -> [a] -> b) -> c -> b -> b iter [] next i done = done iter (x:xs) next i done = next i x (Step iter) xs count :: Int -> Char -> Step Char Int Int -> String -> Int count i x step xs = unStep step xs count (i+1) (i+1) test :: String -> Int test xs = iter xs count 0 0 2- specialize iter for next = count newtype Step a c b = Step { unStep :: [a] -> (c -> a -> Step a c b -> [a] -> b) -> c -> b -> b } iter :: [a] -> (c -> a -> Step a c b -> [a] -> b) -> c -> b -> b iter' :: String -> Int -> Int -> Int iter [] next i done = done iter (x:xs) next i done = next i x (Step iter) xs iter' [] i done = done iter' (x:xs) i done = count i x (Step iter) xs count :: Int -> Char -> Step Char Int Int -> String -> Int count i x step xs = unStep step xs count (i+1) (i+1) test :: String -> Int test xs = iter' xs 0 0 3- specialize count for step = iter (and use the specialized iter') newtype Step a c b = Step { unStep :: [a] -> (c -> a -> Step a c b -> [a] -> b) -> c -> b -> b } iter :: [a] -> (c -> a -> Step a c b -> [a] -> b) -> c -> b -> b iter' :: String -> Int -> Int -> Int iter [] next i done = done iter (x:xs) next i done = next i x (Step iter) xs iter' [] i done = done iter' (x:xs) i done = count' i x xs count :: Int -> Char -> Step Char Int Int -> String -> Int count' :: Int -> Char -> String -> Int count i x step xs = unStep step xs count (i+1) (i+1) count' i x xs = iter' xs (i+1) (i+1) test :: String -> Int test xs = iter' xs 0 0 (iter and count are no longer used and can be dropped at this point) 4- inline count' iter' :: String -> Int -> Int -> Int iter' [] i done = done iter' (x:xs) i done = iter' xs (i+1) (i+1) count' :: Int -> Char -> String -> Int count' i x xs = iter' xs (i+1) (i+1) test :: String -> Int test xs = iter' xs 0 0 (count' is no longer used and can be dropped at this point) 5- eliminate the done argument as it is always the same as the i argument iter'' :: String -> Int -> Int iter'' [] i = i iter'' (x:xs) i = iter'' xs (i+1) test :: String -> Int test xs = iter'' xs 0 Cheers! -Tyson