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haskell - 无法在 haskell 中使用 GADT 实现矩阵和向量

问题描述

我是 Haskell 的初学者。我试图在 haskell 中实现矩阵和向量。我希望这些支持 Num 类型类中的所有类型，但不支持其他类型，所以我尝试使用 GADT，但我遇到了错误，你能解释为什么我会遇到错误。或者我可以不用这种方式获得相同的功能。

我的代码：

{-# LANGUAGE GADTs #-}

class LStruct s where
  scale :: (Num a) => a -> s -> s
  add :: s -> s -> s


data Vector a where
  Vector :: (Num a) => a -> Vector [a]

data Matrix a where
  Matrix :: (Num a) => a -> Matrix [a]


instance LStruct (Vector a) where
  scale a (Vector xs) = Vector (map (*a) xs)
  add (Vector xs) (Vector ys) = Vector (zipWith (+) xs ys)

错误：

[-Wdeferred-type-errors]
• Could not deduce: a2 ~ [a1]
  from the context: (a ~ [a2], Num a2)
    bound by a pattern with constructor:
               Vector :: forall a. Num a => a -> Vector [a],
             in an equation for ‘scale’
    at /home/d4rk0wl/Playground/hs/Mask/.stack-work/intero/intero10129CJD-TEMP.hs:20:12-20
  ‘a2’ is a rigid type variable bound by
    a pattern with constructor:
      Vector :: forall a. Num a => a -> Vector [a],
    in an equation for ‘scale’
    at /home/d4rk0wl/Playground/hs/Mask/.stack-work/intero/intero10129CJD-TEMP.hs:20:12-20
• In the second argument of ‘map’, namely ‘xs’
  In the first argument of ‘Vector’, namely ‘(map (* a) xs)’
  In the expression: Vector (map (* a) xs)
• Relevant bindings include
    xs :: a2
      (bound at /home/d4rk0wl/Playground/hs/Mask/.stack-work/intero/intero10129CJD-TEMP.hs:20:19)
    a :: a1
      (bound at /home/d4rk0wl/Playground/hs/Mask/.stack-work/intero/intero10129CJD-TEMP.hs:20:9)
    scale :: a1 -> Vector a -> Vector a
      (bound at /home/d4rk0wl/Playground/hs/Mask/.stack-work/intero/intero10129CJD-TEMP.hs:20:3)

[-Wdeferred-type-errors]
• Could not deduce: a2 ~ [c0]
  from the context: (a ~ [a1], Num a1)
    bound by a pattern with constructor:
               Vector :: forall a. Num a => a -> Vector [a],
             in an equation for ‘add’
    at /home/d4rk0wl/Playground/hs/Mask/.stack-work/intero/intero10129CJD-TEMP.hs:21:8-16
  or from: (a ~ [a2], Num a2)
    bound by a pattern with constructor:
               Vector :: forall a. Num a => a -> Vector [a],
             in an equation for ‘add’
    at /home/d4rk0wl/Playground/hs/Mask/.stack-work/intero/intero10129CJD-       TEMP.hs:21:20-28
  ‘a2’ is a rigid type variable bound by
    a pattern with constructor:
      Vector :: forall a. Num a => a -> Vector [a],
    in an equation for ‘add’
    at /home/d4rk0wl/Playground/hs/Mask/.stack-work/intero/intero10129CJD-TEMP.hs:21:20-28
  Expected type: [c0]
    Actual type: a1
• In the second argument of ‘zipWith’, namely ‘xs’
  In the first argument of ‘Vector’, namely ‘(zipWith (+) xs ys)’
  In the expression: Vector (zipWith (+) xs ys)
• Relevant bindings include
    ys :: a2
      (bound at /home/d4rk0wl/Playground/hs/Mask/.stack-work/intero/intero10129CJD-TEMP.hs:21:27)

标签： haskellvector

解决方案

我想你打算写：

data Vector a where
    Vector :: (Num a) => [a] -> Vector a

“可以从 a 的列表构造一个 Vector 或 a ”

您问题中的版本需要一个a，并且您得到的错误是关于 GHC 试图映射该单个值。

更广泛地说，我建议将Num约束从构造函数移到使用Vector. 以这种方式编写的代码似乎可以从 GHC 获得更好的错误消息，并且更易于使用。我不确定这是否有深层原因，或者这只是 GHC 开发人员在哪里花费时间的问题。

所以实例声明可能变成：

instance Num a => LStruct (Vector a) where

并且实例主体可以保持不变。

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