2

我正在阅读 Real World Haskell，尝试使用 ghc online解决Ch3、Q10问题。

到目前为止，我有以下代码：

data Direction point = MyLeft point | MyRight point | Straight deriving (Show)

getDirectionFromTriple :: Direction p -> Direction p -> Direction p -> Direction p
getDirectionFromTriple p1 p2 p3 
  | (length . filter (== MyLeft) [p1, p2, p3]) > 1 = MyLeft p3
  | (length . filter (== MyRight) [p1, p2, p3]) > 1 = MyRight p3
  | otherwise = Straight

尝试编译此代码时收到以下错误（仅发布部分，多次弹出相同的错误）：

[1 of 1] Compiling Main             ( jdoodle.hs, jdoodle.o )


jdoodle.hs:17:15: error:
    * Couldn't match expected type `a0 -> t0 a1'
                  with actual type `[point0 -> Direction point0]'
    * Possible cause: `filter' is applied to too many arguments
      In the second argument of `(.)', namely
        `filter (== MyLeft) [p1, p2, p3]'
      In the first argument of `(>)', namely
        `(length . filter (== MyLeft) [p1, p2, p3])'
      In the expression: (length . filter (== MyLeft) [p1, p2, p3]) > 2
   |
17 |   | (length . filter (== MyLeft) [p1, p2, p3]) > 2 = MyLeft p3
   |               ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

jdoodle.hs:17:35: error:
    * Couldn't match expected type `point0 -> Direction point0'
                  with actual type `Direction p'
    * In the expression: p1
      In the second argument of `filter', namely `[p1, p2, p3]'
      In the second argument of `(.)', namely
        `filter (== MyLeft) [p1, p2, p3]'
    * Relevant bindings include
        p3 :: Direction p (bound at jdoodle.hs:16:30)
        p2 :: Direction p (bound at jdoodle.hs:16:27)
        p1 :: Direction p (bound at jdoodle.hs:16:24)
        getDirectionFromTriple :: Direction p
                                  -> Direction p -> Direction p -> Direction p
          (bound at jdoodle.hs:16:1)
   |
17 |   | (length . filter (== MyLeft) [p1, p2, p3]) > 2 = MyLeft p3
   |                                   ^^

jdoodle.hs:17:39: error:
    * Couldn't match expected type `point0 -> Direction point0'
                  with actual type `Direction p'
    * In the expression: p2
      In the second argument of `filter', namely `[p1, p2, p3]'
      In the second argument of `(.)', namely
        `filter (== MyLeft) [p1, p2, p3]'
    * Relevant bindings include
        p3 :: Direction p (bound at jdoodle.hs:16:30)
        p2 :: Direction p (bound at jdoodle.hs:16:27)
        p1 :: Direction p (bound at jdoodle.hs:16:24)
        getDirectionFromTriple :: Direction p
                                  -> Direction p -> Direction p -> Direction p
          (bound at jdoodle.hs:16:1)
   |
17 |   | (length . filter (== MyLeft) [p1, p2, p3]) > 2 = MyLeft p3
   |                                       ^^

我将不胜感激有关如何修复我的代码的建议或有关从三点确定主要方向的更简洁解决方案的建议。

4

2 回答 2

2

您可以创建函数来告诉您它是右还是左，而不是使用==

import Data.List 

data Direction point = MyLeft point | MyRight point | Straight deriving (Show)

getDirectionFromTriple :: Direction p -> Direction p -> Direction p -> Direction p
getDirectionFromTriple p1 p2 p3 
  | isDirection isLeft  [p1, p2, p3]  = MyLeft (getValue p3)
  | isDirection isRight [p1, p2, p3]  = MyRight (getValue p3)
  | otherwise = Straight

isDirection :: (Direction p -> Bool) -> [Direction p] -> Bool
isDirection f ps = (length . filter f) ps > 1

getValue (MyLeft a) = a
getValue (MyRight a) = a
getValue Straight  = error "No value in Straight"

isLeft  (MyLeft _)  = True
isLeft   _          = False
isRight (MyRight _) = True
isRight  _          = False

main = do
  putStrLn $ show $ getDirectionFromTriple (MyRight 2) Straight (MyLeft 1)
  putStrLn $ show $ getDirectionFromTriple (MyRight 2) (MyRight 3) (MyLeft 1)
  putStrLn $ show $ getDirectionFromTriple (MyLeft 1) (MyLeft 1) (MyRight 2)
于 2019-10-01T17:02:12.523 回答
2

与最近的问题相同，为什么odd.fst 不适用于过滤器功能？. 当你写

length . filter f xs

这被解析为

length . (filter f xs)

这显然是不对的：filter f xs是一个列表，而不是一个函数，所以用 . 组合它是没有意义的length。相反，你可以写

(length . filter f) xs

虽然更流行的拼写方式是

length . filter f $ xs
于 2019-10-01T15:25:14.987 回答