Haskell-guard内部案例陈述
我正在浏览这本书,在第8章中有一段代码,看起来像这样Haskell-guard内部案例陈述,haskell,pattern-guards,Haskell,Pattern Guards,我正在浏览这本书,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_ m
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
这很有效。但是,我想将if/else块转换为如下的保护语句:
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
这不能编译。Haskell中警卫的使用似乎非常严格/不直观。有没有一个确切的来源,我可以阅读,告诉我在哪些地方可以使用警卫 有两个地方是允许的:函数定义和
case
表达式。在这两种情况下,防护装置都出现在图案之后和主体之前,因此您可以像往常一样在函数中使用=
,在案例中使用->
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
仅匹配包装值也是正的Just
我认为最终参考是,特别是§3.13 Case表达式和§4.4.3函数和模式绑定,它们描述了保护的语法并指定了允许它们的位置
在代码中,您需要:
Just (state, code)
| state == Taken -> Left "LockerNumber already taken!"
| otherwise -> Right code
这也可以仅通过模式来表达:
Just (Taken, _) -> Left "LockerNumber already taken!"
Just (_, code) -> Right code
回答得好!尤其是行防护只是模式的约束