On Thu, 20 Dec 2007, Stefan O'Rear wrote:
On Thu, Dec 20, 2007 at 11:39:42PM -0500, Ronald Guida wrote:
data PZero = PZero deriving (Show) data PSucc a = PSucc a deriving (Show)
type P1 = PSucc PZero type P2 = PSucc P1 type P3 = PSucc P2 -- etc
...
Now here's the puzzle. I want to create a function "vecLength" that accepts a vector and returns its length. The catch is that I want to calculate the length based on the /type/ of the vector, without looking at the number of elements in the list.
So I started by defining a class that allows me to convert a Peano number to an integer. I couldn't figure out how to define a function that converts the type directly to an integer, so I am using a two-step process. Given a Peano type /t/, I would use the expression "pToInt (pGetValue :: t)".
class Peano t where pGetValue :: t pToInt :: t -> Int
instance Peano PZero where pGetValue = PZero pToInt _ = 0
instance (Peano t) => Peano (PSucc t) where pGetValue = PSucc pGetValue pToInt (PSucc a) = 1 + pToInt a
Why the two methods pGetValue and pToInt? I thought that pGetValue should have signature pGetValue :: t -> Peano and that pToInt can convert the result from pGetValue to Int.
Finally, I tried to define vecLength, but I am getting an error.
vecLength :: (Peano s) => Vec s t -> Int vecLength _ = pToInt (pGetValue :: s)
1. pGetValue is unneccessary, undefined :: s will work just as well. This is a fairly standard approach; the precision values in Floating, bitSize :: Bits a => a -> Int, and sizeOf :: Storable a => a -> Int all work this way.
What about functions of type bitSize :: Bits a => (a, Int) This would also make explicit, that you don't have to provide a value of type 'a'. 'sizeOf' might still return different size depending on the value, right? E.g. when converting a sum data type to a (C++) object.