你如何对 Haskell 中的类型类和数据类型进行递归？

Question

I am working on a type class for fraction, vector, and matrix arithmetic (ie add, sub, mul) but can't quite get the vector instance because I don't know how to work with the recursive nature of the function.

here's the code:

class MathObject a where
    add :: a -> a -> a
    sub :: a -> a -> a
    mul :: a -> a -> a


type Vector = [Int]


data Vec = Vec Vector deriving (Show, Eq)


instance MathObject Vec where 
    add (Vec v1) (Vec v2) = addVecs (Vec v1) (Vec v2)
    sub (Vec v1) (Vec v2) = subVecs (Vec v1) (Vec v2)
    mul (Vec v1) (Vec v2) = mulVecs (Vec v1) (Vec v2)

addVecs :: Vec -> Vec -> [Int]
addVecs (Vec []) (Vec vec2)= (Vec [])
addVecs (Vec vec) (Vec []) = (Vec [])
addVecs (Vec (x:xs)) (Vec (y:ys)) = e1+e2:rest where
  e1 = x
  e2 = y
  rest = addVecs (Vec xs) (Vec ys)

subVecs :: Vec -> Vec -> [Int]
subVecs (Vec []) (Vec v2) = (Vec [])
subVecs (Vec (x:xs)) (Vec (y:ys)) = x-y:rest where
  rest = subVecs (Vec xs) (Vec ys)

mulVecs :: Vec -> Vec -> [Vec]
mulVecs (Vec []) (Vec vec2) = (Vec [])
mulVecs (Vec (x:xs)) (Vec (y:ys)) = e1 * e2:rest where
  e1 = x
  e2 = y
  rest = mulVecs (Vec xs) (Vec ys)

I made the fraction instance and it works so I have some basic understanding of how type classes work but I just don't know how to deal with a recursive type.

Answer 1

I would say, first, you have confused yourself with too many different things named something that sounds like "vector". Why does the Vec type have a constructor named Vec and fields of type Vector? What's the difference really between a Vec and a Vector? If Vector is an alias for [Int], why do you use ordinary [Int] to represent vectors sometimes? You don't include the compiler errors you get, but I can see clearly that at least some of them will arise due to type mismatches between these names.

So the first thing to do is simplify it: just have one Vector type, which is a newtype (or data if you haven't gotten to newtypes yet) wrapper around [Int] :

newtype Vector = Vector [Int]

Now you always know whether you're working with a Vector or an [Int], and can convert between the two using the Vector constructor.

My next observation would be that addVecs and friends clearly have the wrong type: why do they take two Vector inputs and return an [Int] ? They should stick with one type, preferably Vector . This is the cause of one of your type errors: add needs to be of type a -> a -> a , but you've defined add to be addVecs , which has type Vec -> Vec -> [Int] - that doesn't fit! So let's instead write

addVecs :: Vector -> Vector -> Vector
-- ...

Now, the implementation is right given the poor type you chose for it. But to work for the better type, we'd have to add a Vector wrapper around a+b before consing it to the result. I won't show that implementation, though, because there's a much better one available. Instead of doing the recursion yourself, use one of Haskell's many flexible tools for working with lists: zipWith .

addVecs (Vector v1) (Vector v2) = Vector $ zipWith (+) v1 v2

This does exactly the same thing as your implementation, but all the tedious wrapping and unwrapping is done only once, and recursive list processing is completely abstracted away by zipWith .

You could do the same thing for subVecs and mulVecs , but you would quickly notice there's some duplication here. It would be better to extract that out into a function, so that those three functions can all share the common code:

pointwise :: (Int -> Int -> Int) -> (Vector -> Vector -> Vector)
pointwise f (Vector v1) (Vector v2) = Vector $ zipWith f v1 v2

addVecs = pointwise (+)
subVecs = pointwise (-)
mulVecs = pointwise (*)

At this point you don't even really need definitions for addVecs and friends - you could easily inline them into the MathObject instance:

instance MathObject Vector where
  add = pointwise (+)
  sub = pointwise (-)
  mul = pointwise (*)