Parsing 如何解析中缀而不是带有Haskell的前缀？

我需要帮助这个程序，我正试图写在哈斯克尔。我写了大部分，下面是我基本上要做的：

当我写作时

解析“a+b”

在终端中，我希望将其作为输出：

加上（单词“a”）（单词“b”）

当我写作时

解析“a-2*b+c”

在终端中，我希望将其作为输出：

减（单词“a”）（加（单词“b”）（单词“c”）

到目前为止，我的代码是：

data Ast
    = Word String
    | Num Int
    | Mult Ast Ast
    | Plus Ast Ast
    | Minus Ast Ast
    deriving (Eq, Show)

tokenize :: [Char] -> [String]
tokenize [] = []
tokenize (' ' : s) = tokenize s
tokenize ('+' : s) = "+" : tokenize s
tokenize ('*' : s) = "*" : tokenize s
tokenize (c : s)
  | isDigit c =
    let (cs, s') = collectWhile isDigit s
     in (c : cs) : tokenize s'
  | isAlpha c =
    let (cs, s') = collectWhile isAlpha s
     in (c : cs) : tokenize s'
  | otherwise = error ("unexpected character " ++ show c)

collectWhile :: (Char -> Bool) -> String -> (String, String)
collectWhile p s = (takeWhile p s, dropWhile p s)

isDigit, isAlpha :: Char -> Bool
isDigit c = c `elem` ['0' .. '9']
isAlpha c = c `elem` ['a' .. 'z'] ++ ['A' .. 'Z']

parseU :: [String] -> (Ast, [String])
parseU ("+" : s0) =
  let (e1, s1) = parseU s0
      (e2, s2) = parseU s1
   in (Plus e1 e2, s2)
parseU ("*" : s0) =
  let (e1, s1) = parseU s0
      (e2, s2) = parseU s1
   in (Mult e1 e2, s2)
parseU (t : ts)
  | isNumToken t = (Num (read t), ts)
  | isWordToken t = (Word t, ts)
  | otherwise = error ("unrecognized token " ++ show t)
parseU [] = error "unexpected end of input"

isNumToken, isWordToken :: String -> Bool
isNumToken xs = takeWhile isDigit xs == xs
isWordToken xs = takeWhile isAlpha xs == xs

parse :: String -> Ast
parse s =
  case parseU (tokenize s) of
    (e, []) -> e
    (_, t : _) -> error ("unexpected token " ++ show t)

inn :: Ast -> String
inn (Plus x y) = innP x ++ " + " ++ innP y
inn (Mult x y) = innP x ++ " * " ++ innP y
inn ast = innP ast

innP :: Ast -> String
innP (Num n) = show n
innP (Plus x y) = "(" ++ innP x ++ " + " ++ innP y ++ ")"
innP (Mult x y) = "(" ++ innP x ++ " * " ++ innP y ++ ")"
innP (Word w) = w -- 

innfiks :: String -> String
innfiks s = inn (parse s)

现在，我在终端上写的文本发布时出错，但当我这样写时：

解析“+AB”

我得到了正确的输出：

加上（单词“a”）（单词“b”）

我知道我必须更改代码，以便它接受我发送给此表单上解析函数的内容：

值运算符值

而不是在本表格上：

运算符值

---

但是我正在努力找出我必须做的改变。

要处理具有优先级的中缀运算符，一种方法是引入一系列与优先级对应的解析函数。因此，如果您有“因子”，可以将其相乘以创建“术语”，可以将其相加或相减以创建“表达式”，那么您将需要为每个级别创建解析器函数。解析“因子”（即单词或数字）很容易，因为您已经编写了以下代码：

parseFactor :: [String] -> (Ast, [String])
parseFactor (t : ts)
  | isNumToken t = (Num (read t), ts)
  | isWordToken t = (Word t, ts)
  | otherwise = error ("unrecognized token " ++ show t)
parseFactor [] = error "unexpected end of input"

解析一个术语更为棘手。您希望首先解析一个因子，然后可选地解析一个

*

，后跟另一个因子，然后将其视为一个术语，进一步可选地与另一个因子相乘，依此类推。以下是一种方法：

parseTerm :: [String] -> (Ast, [String])
parseTerm ts
  =  let (f1, ts1) = parseFactor ts     -- parse first factor
     in  go f1 ts1
  where go acc ("*":ts2)                -- add a factor to an accumulating term
          = let (f2, ts3) = parseFactor ts2
            in go (Mult acc f2) ts3
        go acc rest = (acc, rest)       -- no more factors: return the term

如果需要，请尝试编写一个类似的

parsexpr

来解析由

+

字符分隔的术语（现在跳过减法），并在以下情况下对其进行测试：

parseExpr (tokenize "2 + 3 * 6 + 4 * 8 * 12 + 1")

对于破坏者，这里有一个版本可以同时处理

+

和

-

，但是请注意，您的标记器还不能正确处理减法，因此您必须首先解决这个问题

parseExpr :: [String] -> (Ast, [String])
parseExpr ts
  =  let (f1, ts1) = parseTerm ts
     in  go f1 ts1
  where go acc (op:ts2)
          | op == "+" || op == "-"
          = let (f2, ts3) = parseTerm ts2
            in go ((astOp op) acc f2) ts3
        go acc rest = (acc, rest)
        astOp "+" = Plus
        astOp "-" = Minus

有了它，您可以将

parse

指向正确的解析器：

parse :: String -> Ast
parse s =
  case parseExpr (tokenize s) of
    (e, []) -> e
    (_, t : _) -> error ("unexpected token " ++ show t)

你的例子应该有用：

λ> parse "a - 2 * b + c"
Plus (Minus (Word "a") (Mult (Num 2) (Word "b"))) (Word "c")

请注意，这与您所希望的输出略有不同，但这种顺序对于左关联运算符是正确的（这对于正确处理

-

）。也就是说，您想要：

5 - 4 + 1

解析为：

(5 - 4) + 1  -- i.e., (Plus (Minus (Num 5) (Num 4)) (Num 1))

以便评估人员计算2的正确答案。如果将其解析为：

5 - (4 + 1)  -- i.e., (Minus (Num 5) (Plus (Num 4) (Num 1)))

您的评估者将计算错误的答案0

但是，如果您真的想使用正确的关联运算符进行解析，请参见下文

左关联运算符的完整修改代码：

data Ast
    = Word String
    | Num Int
    | Mult Ast Ast
    | Plus Ast Ast
    | Minus Ast Ast
    deriving (Eq, Show)

tokenize :: [Char] -> [String]
tokenize [] = []
tokenize (' ' : s) = tokenize s
tokenize ('-' : s) = "-" : tokenize s
tokenize ('+' : s) = "+" : tokenize s
tokenize ('*' : s) = "*" : tokenize s
tokenize (c : s)
  | isDigit c =
    let (cs, s') = collectWhile isDigit s
     in (c : cs) : tokenize s'
  | isAlpha c =
    let (cs, s') = collectWhile isAlpha s
     in (c : cs) : tokenize s'
  | otherwise = error ("unexpected character " ++ show c)

collectWhile :: (Char -> Bool) -> String -> (String, String)
collectWhile p s = (takeWhile p s, dropWhile p s)

isDigit, isAlpha :: Char -> Bool
isDigit c = c `elem` ['0' .. '9']
isAlpha c = c `elem` ['a' .. 'z'] ++ ['A' .. 'Z']

parseFactor :: [String] -> (Ast, [String])
parseFactor (t : ts)
  | isNumToken t = (Num (read t), ts)
  | isWordToken t = (Word t, ts)
  | otherwise = error ("unrecognized token " ++ show t)
parseFactor [] = error "unexpected end of input"

parseTerm :: [String] -> (Ast, [String])
parseTerm ts
  =  let (f1, ts1) = parseFactor ts
     in  go f1 ts1
  where go acc ("*":ts2)
          = let (f2, ts3) = parseFactor ts2
            in go (Mult acc f2) ts3
        go acc rest = (acc, rest)

parseExpr :: [String] -> (Ast, [String])
parseExpr ts
  =  let (f1, ts1) = parseTerm ts
     in  go f1 ts1
  where go acc (op:ts2)
          | op == "+" || op == "-"
          = let (f2, ts3) = parseTerm ts2
            in go ((astOp op) acc f2) ts3
        go acc rest = (acc, rest)
        astOp "+" = Plus
        astOp "-" = Minus

isNumToken, isWordToken :: String -> Bool
isNumToken xs = takeWhile isDigit xs == xs
isWordToken xs = takeWhile isAlpha xs == xs

parse :: String -> Ast
parse s =
  case parseExpr (tokenize s) of
    (e, []) -> e
    (_, t : _) -> error ("unexpected token " ++ show t)

对于右关联运算符，请修改以下定义：

parseTerm :: [String] -> (Ast, [String])
parseTerm ts
  =  let (fct, ts1) = parseFactor ts
     in  case ts1 of
           "*":ts2 -> let (trm, rest) = parseTerm ts2
                      in  (Mult fct trm, rest)
           _       -> (fct, ts1)

parseExpr :: [String] -> (Ast, [String])
parseExpr ts
  =  let (trm, ts1) = parseTerm ts
     in  case ts1 of
           op:ts2 | op == "+" || op == "-"
                   -> let (expr, rest) = parseExpr ts2
                      in  (astOp op trm expr, rest)
           _       -> (trm, ts1)
  where astOp "+" = Plus
        astOp "-" = Minus*

---

这里有一个解析器。关于如何进一步解析中缀符号，我们可以找到大量的资料。正如你在回答中所说的，外部是加号，内部是负号。有没有办法改变这一点？我尝试了很多方法，但似乎最后一个表达式总是排在第一位。我已经更新了我的答案来展示如何做到这一点。但是请注意，如果您想要一个以“正常方式”解析数学表达式的解析器，实际上您需要外部加上，内部减下，正如我在上面所解释的。非常感谢：-）