haskell 正在解析上下文无关语法以与字符串进行比较

Question

我试图表示一个上下文无关的语法，我想解析它以将字符串与 Prod "A" [NT "B", NT "C"] 的 "A" 进行比较，但我不知道如何，有人可以请帮帮我？

data Symbol a b = T a | NT b 
deriving (Eq,Show)

data Produktion a b = Prod b [Symbol a b]
deriving (Eq,Show)

type Produktionen a b = [Produktion a b]   

liste11 = [Prod "A" [NT "B", NT "C"]
      ,Prod "B" [NT "D", NT "E"]
      ,Prod "D" [T "d"]
      ] --

Answer 1

有更简单的方法，但可以使用原始方法

fromNT :: (Symbol a b) -> b
fromNT (NT b) = b
fromNT (T a) = undefined

extractNT :: (Eq b) => b -> [Produktion a b] -> [[b]]   
extractNT _ [] = [[]]
extractNT b (p:ps)  | is p = (map fromNT (symbol p)) : extractNT b ps
                    | otherwise = []          
                    where is (Prod x xs) = x==b
                          symbol (Prod _ xs) = xs

现在你可以写了

> extractNT "A" liste11                         
[["B","C"]]

更新

我认为它不适用于泛型类型，但适用于 String String

value :: (Symbol String String) -> String
value (T a) = a
value (NT b) = b

其他访问器仍然可以是通用的

symbol :: (Produktion a b) -> [Symbol a b]
symbol (Prod _ ss) = ss

tag :: (Produktion a b) -> b
tag (Prod b _) = b

你现在可以说

> map (map value) $ map symbol $ filter (\p -> tag p=="D") liste11
[["d"]]

> map (map value) $ map symbol $ filter (\p -> tag p=="A") liste11
[["B","C"]]

haskell 正在解析上下文无关语法以与字符串进行比较

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