At 4:27 AM +0000 2003/11/06, Patty Fong wrote:
data Music = Note Pitch Octave Duration | Silence Duration | PlayerPar Music Music | PlayerSeq Music Music | Tempo (Ratio Int) Music data Pitch = Cf | C | Cs type Octave = Int type Duration = Ratio Int
foldMusic :: (Pitch -> Octave -> Duration -> a) -> (Duration -> a) -> (a -> a -> a) -> (a -> a -> a) -> (Ratio Int -> a -> a) -> Music -> a
foldMusic n _ _ _ _ (Note pitch octive duration) = n pitch octive duration foldMusic _ s _ _ _ (Silence duration) = s duration foldMusic n s p1 p2 t (PlayerPar partOne partTwo) = p1 (foldMusic n s p1 p2 t partOne)(foldMusic n s p1 p2 t partTwo) foldMusic n s p1 p2 t (PlayerPar partA partB) = p2 (foldMusic n s p1 p2 t partA)(foldMusic n s p1 p2 t partB) foldMusic n s p1 p2 t (Tempo rate part) = t rate (foldMusic n s p1 p2 t part)
I understand that when i use the foldMusic function i need to pass it 5 parameters.
Actually, 6 parameters are required before the function invocation is complete.
given the type signiature, why can i pass (+) as a parameter for p1 but not for n, what determines what can be passed as a parameter, because they all have the return type a??
A match between two function types requires not only that the return types match, but also that there are the same number of parameters and that the parameter types match.
I attempted to create a function that utilises the foldMusic function that counts the number of notes:
count_notes :: Music -> Integer count_notes = foldMusic (\_-> \_ -> \_ -> 1) (\_ -> 0) (+) (+) (\_ -> \_ -> 0)
it appears to work, i think. Yet i'm still not certain of how it does so.
Very close. The function for "Tempo" needs fixing. Try the following (untested code): count_notes = foldMusic (\_ _ _ -> 1) (\_ -> 0) (+) (+) (\_ m -> m)
Is there anyway to represent other fold functions in a tree like representation as foldr (+) 0 would appear as such? + 1 \ + 2 \ + 3 \ 0
Yes, but it's too late for me to draw the more complicated diagram that would correspond to your Music example. Dean P.S. At 3:41 PM +1100 2003/11/06, Thomas L. Bevan wrote:
what you have written is not a fold. A fold operates over a list. There is no list in your code, only some sort of tree structure.
To me, a "fold" operates over any recursively defined (aka "inductive") data type. With this more general definition, what Patty has written is most certainly a fold. In fact, "count_notes" above corresponds directly to Thomas's "countNotes":
countNotes Silence _ = 0 countNotes Note_ = 1 countNotes PlayerPar m1 m2 = (countNotes m1) + (countNotes m2) countNotes PlayerSeq m1 m2 = (countNotes m1) + (countNotes m2) countNotes Tempo _ m = countNotes m