Haskell中的非穷举函数

我用Haskell编写了一个函数：

recognize :: [GrammarRule] -> State -> [String] -> Bool
recognize [] x [s] = False
recognize gr x [s] = if elem True (map (\state -> elem state (takeSteps gr x [s]))(enders gr)) then True else False

这并非详尽无遗。识别功能有什么问题？ 非常感谢

--the function takesteps is here: 
takeSteps :: [GrammarRule] -> State -> [String] -> [State]
takeSteps g s []     = [s]
takeSteps g s (w:ws) = concatMap (\nextState -> takeSteps g nextState ws) (successors g s w)

--the function enders is here:
enders :: [GrammarRule] -> [State]
enders [] = []
enders (r:rs) =
    let resultFromRest = enders rs in
    case r of
    End x -> x : resultFromRest
    Step x w' y -> resultFromRest

---

[s]

只匹配一个元素的列表，

s

如果要匹配任何列表，请使用简单标识符，

l

，例如：

recognize [] x [s] = False
recognize gr x l = if elem True (map (\state -> elem state (takeSteps gr x l))(enders gr)) then True else False
                                               -- note the usage of l here ^

---

避免

如果x为真，否则为假

，这是多余的。只需使用

x

。还可以使用

-Wall

打开警告，以便尽早捕获这些错误。