Question

我想定义几个版本的东西，但是使用不同的类型来增强程序中的类型安全性。例如，我有几种类型的双变量值，我希望它们是Num的实例，但都应该是不同的类型。所以我所做的是创建一个带有一个类型变量的newtype并根据它声明新类型。但是，我觉得有点烦人，我现在必须一直使用两个构造函数。有没有办法解决？

{-# LANGUAGE GeneralizedNewtypeDeriving #-}

newtype Bivar t = Bivar (t,t) deriving (Show, Eq)

instance (Num t) => Num (Bivar t) where 
  (+) (Bivar (x1,y1)) (Bivar (x2,y2)) = Bivar (x1+x2, y1+y2)
  (-) (Bivar (x1,y1)) (Bivar (x2,y2)) = Bivar (x1-x2, y1-y2)
  (*) (Bivar (x1,y1)) (Bivar (x2,y2)) = Bivar (x1*x2, y1*y2)
  abs (Bivar (x1,y1)) = Bivar (abs x1, abs y1)
  fromInteger i = Bivar (fromInteger i, fromInteger i)
  signum (Bivar (x1,y1)) = Bivar (signum x1, signum y1)

newtype BivarNode = BivarNode (Bivar Int) deriving (Show, Eq, Num)
newtype BivarVal = BivarVal (Bivar Double) deriving (Show, Eq, Num)
newtype HBivarVal = HBivarVal (Bivar Double) deriving (Show, Eq, Num)

-- This is annoying:
a1 = BivarVal (Bivar (1.0, 2.0))
a2 = HBivarVal (Bivar (1.0, 2.0))
b = BivarNode (Bivar (1,2))

-- is there a way so that I can write it this way?
aa1 = BivarVal (1.0, 2.0)
aa2 = HBivarVal (1.0, 2.0)
bb = BivarNode (1,2)

谢谢你！

修改

为了扩展原始问题，我还想在模式匹配上使用类型名称，类似于

myFunction :: HBivarVal -> Double
myFunction (HBivarVal (Bivar (x,y))) = x

也可以吗？

Answer 1

为什么不将所有作为新类型，甚至是Bivar？

instance (Num a, Num b) => Num (a, b) where
    (a, b) + (a', b') = (a+a', b+b')
    (a, b) * (a', b') = (a*a', b*b')
    (a, b) - (a', b') = (a-a', b-b')
    fromInteger i = (fromInteger i, fromInteger i)
    abs    (a, b) = (abs    a, abs    b)
    signum (a, b) = (signum a, signum b)

newtype Bivar t   = Bivar     (t     , t     ) deriving (Show, Eq, Num)
newtype BivarNode = BivarNode (Int   , Int   ) deriving (Show, Eq, Num)
newtype BivarVal  = BivarVal  (Double, Double) deriving (Show, Eq, Num)
newtype HBivarVal = HBivarVal (Double, Double) deriving (Show, Eq, Num)

Answer 2

是的，您可以定义辅助函数：

bivarVal = BivarVal . Bivar
hBivarVal = HBivarVal . Bivar
bivarNode = BivarNode . Bivar

Haskell：类型安全的同义词的新类型：使用两个构造函数的任何方式？

2 个答案: