Question

如果我从GHCi终端执行myZipWith (+) (Point [1,2,3]) (Point [4,5,6])，一切正常，但是如果我尝试从一个简单的函数执行它，就会给我错误。

为什么即使代码相同，直接从终端执行的内容也不同？

这是我的代码：

data Point p = Point [p]
deriving (Show)

wtf :: [p]
wtf = myZipWith (+) (Point [1,2,3]) (Point [4,5,6])

myZip :: Point p -> Point p -> [(p, p)]
myZip (Point []) _ = []
myZip _ (Point []) = []
myZip (Point (a:as)) (Point (b:bs)) = (a, b) : myZip (Point as) (Point bs)


myZipWith :: (p -> p -> p) -> Point p -> Point p -> [p]
myZipWith f (Point p1) (Point p2) = [ f (fst x) (snd x) | x <- (myZip (Point p1) (Point p2)) ]

错误代码：

No instance for (Num p) arising from a use of `+'
      Possible fix:
        add (Num p) to the context of
          the type signature for:
            wtf :: forall p. [p]
    * In the first argument of `myZipWith', namely `(+)'
      In the expression:
        myZipWith (+) (Point [1, 2, 3]) (Point [4, 5, 6])
      In an equation for `wtf':
          wtf = myZipWith (+) (Point [1, 2, 3]) (Point [4, 5, 6])

Answer 1

您已定义wtf适用于任何类型的p，而不仅仅是具有Num实例的类型。您需要在类型注释中包含约束。

wtf :: Num p => [p]
wtf = myZipWith (+) (Point [1,2,3]) (Point [4,5,6])

这是建议的可能解决方法所暗示的解决方案（即“将类型Num p添加到类型签名的上下文中”）。 myZip和myZipWith不需要注释，因为它们与Num无关。仅使用wtf的{{1}}的定义需要附加约束。

如果您放弃类型注释，只写了(+) :: Num a => a -> a -> a，那么Haskell就会为列表推断类型wtf = myZipWith (+) (Point [1,2,3]) (Point [4,5,6])。您明确的，过于笼统的类型阻止了推理的发生。

（Num p）不会因使用'+'而引起

1 个答案: