On Wed, May 23, 2012 at 12:51 PM, Dmitriy Matrosov
On 05/22/12 06:18, Brent Yorgey wrote:
However, you should be able to define a general fold for Tape, with type
foldTape :: (a -> [b] -> b) -> Tape a -> b
So, you wrote it correctly :
foldTape :: (a -> [b] -> b) -> Tape a -> b foldTape f (Tape x ts) = f x (map (foldTape f) ts)
This appears desperate since there's no mention of Int anywhere in this function and the function applied stay the same whatever the level : the initially given f parameter. But there's a trick to this, we'll have to treat some of our type variable like functions, here only b can vary freely (a is imposed by the Tape a inputted), so let's see what it looks like if we make it a functional type (with Int parameter):
foldTape :: (a -> [Int -> b] -> (Int -> b)) -> Tape a -> (Int -> b)
much more promising wouldn't you say ? The solution now looks like that :
foldTapeD :: (Monoid m) => (Int -> a -> m) -> Tape a -> m foldTapeD f t = (foldTape go t) 0 where go x fs n = ....
I let you write your solution (if you didn't find before tomorrow evening, I'll give you the answer). You can then call foldTapeD thus :
foldTapeD (\n x -> if n < 2 then [x] else []) testTape
(much nicer than your initial solution, is it not ?) -- Jedaï