Haskell - 守卫案件陈述

Question

我正在通过了解你是一本哈斯克书,在第8章中有一段代码看起来像这样

data LockerState = Taken | Free deriving (Eq, Show)
type Code = String
type LockerMap = Map.Map Int (LockerState, Code)

lookup' :: Int -> LockerMap -> Either String Code
lookup' num_ map_ =
   case (Map.lookup num_ map_) of
      Nothing -> Left $ "LockerNumber doesn't exist!"
      Just (state, code) -> if state == Taken
                              then Left $ "LockerNumber already taken!"
                              else Right $ code

这有效.但是,我想将if/else块转换为保护语句,如下所示:

lookup' :: Int -> LockerMap -> Either String Code
lookup' num_ map_ =
   case (Map.lookup num_ map_) of
      Nothing -> Left $ "LockerNumber doesn't exist!"
      Just (state, code) ->
         | state == Taken = Left $ "LockerNumber already taken!"
         | otherwise = Right $ Code

这不编译.似乎在Haskell中使用防护是非常严格/非直观的.SO Ex1 SO Ex2.是否有一个明确的来源,我可以阅读,告诉我哪些地方我可以使用警卫？

Answer 1

允许使用两个防护:功能定义和case表达式.在两种情况下,后卫出现后一种模式和之前的身体,让你使用=的功能和->在case树枝,像往常一样:

divide x y
  | y == 0 = Nothing
  --------
  | otherwise = Just (x / y)
  -----------

positively mx = case mx of
  Just x | x > 0 -> Just x
         -------
  _ -> Nothing

防护只是模式的约束,因此Just x匹配任何非Nothing值,但Just x | x > 0只匹配Just其包装值也是正值的.

我认为最终的引用是Haskell报告,特别是§3.13案例表达式和§4.4.3函数和模式绑定,它们描述了守卫的语法并指定了它们被允许的位置.

在您的代码中,您需要:

Just (state, code)
  | state == Taken -> Left "LockerNumber already taken!"
  | otherwise -> Right code

仅使用模式也可以表达这一点:

Just (Taken, _) -> Left "LockerNumber already taken!"
Just (_, code) -> Right code