如何更改 Haskell 函数的类型

Question

我有一个函数，其类型为：

mulPoly :: Poly -> Poly -> Poly

但是我想要一个类型：

mulPoly :: (Num a, Eq a) => Poly a -> Poly a -> Poly a

如何转换此 Haskell 代码：

data Poly = [Int] deriving (Show, Eq)

mulPoly :: Poly -> Poly -> Poly
mulPoly (a:as) bs = 
  addPoly (scale a bs) (0 : mulPoly as bs)

此类型：

data Poly a = P [a] deriving (Show, Eq)

mulPoly :: Poly a -> Poly a -> Poly a

这里是函数 addPoly 和 scale 已经在类型 Poly a

上定义

addPoly :: (Num a, Eq a) => Poly a -> Poly a -> Poly a
addPoly (P as) (P bs) = P (inner as bs)
  where
    inner [] ys = ys
    inner xs [] = xs
    inner (x:xs) (y:ys) = (x+y) : inner xs ys

scale :: (Num a, Eq a) => a -> Poly a -> Poly a
scale 0 (P (x:xs)) = P []
scale y (P []) = P []
scale y (P (x:xs)) = P (map (y*) (x:xs))

Answer 1

只需要在适当的地方做一些包装和解包：

mulPoly :: (Num a, Eq a) => Poly a -> Poly a -> Poly a
mulPoly (P []) _ = P []   -- you were missing this case
mulPoly (P (a:as)) bs =
    addPoly (scale a bs) 
            (shift1 $ mulPoly (P as) bs)
  where
    shift1 (P bs) = P (0:bs)

我们只需要注意什么将原始列表作为参数，什么将它们包装在 P 标签下，并相应地调整您的原始代码。

如何更改 Haskell 函数的类型

How can I change the type of my Haskell function

haskell

types

polynomials