问题描述
我需要用Haskell编写的该程序的帮助。我已经写了大部分,这基本上是我要做的:
- 我写的时候
解析“ a + b”
在终端中,我希望将其作为输出:
加号(单词“ a”)(单词“ b”)
- 我写的时候
解析“ a-2 * b + c”
在终端中,我希望将其作为输出:
减号(单词“ a”)(加号(多(数字2)(单词“ b”))(单词“ c”))
到目前为止,我的代码:
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 (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,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,我在终端上发布我写的文本时出错,但是当我这样写时:
解析“ + a b”
我得到正确的输出:
加号(单词“ a”)(单词“ b”)
我知道我必须更改代码,以便它接受我以这种形式发送给parse函数的内容:
值运算符值
而不是这种形式:
运算符值
但是我在努力找出需要做些什么和在哪里进行此更改。
解决方法
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相对应的解析函数。因此,如果您具有可以乘以创建“项”的“因子”,可以将这些“项”加或减以创建“表达式”,则需要为每个级别创建解析器函数。解析“因数”(即单词或数字)非常容易,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"
解析术语比较棘手。您想先解析一个因数,然后解析一个*
,然后再解析另一个因数,然后将其视为要进一步可选地乘以另一个因数的项,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,请尝试编写类似的parseExpr
来解析用+
字符分隔的术语(暂时跳过减法),然后在类似以下内容上进行测试:
parseExpr (tokenize "2 + 3 * 6 + 4 * 8 * 12 + 1")
对于扰流板,这是一个可以处理+
和-
的版本,尽管请注意,您的令牌生成器尚未正确处理减法,因此您必须先对其进行修复。
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
在适当的位置,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)
您的示例应该可以工作:
λ> parse "a - 2 * b + c"
Plus (Minus (Word "a") (Mult (Num 2) (Word "b"))) (Word "c")
请注意,这与您想要的输出略有不同,但是此顺序对于左关联运算符是正确的(这对于正确处理-
很重要)。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))
,以便评估者计算出正确的答案2。如果将其解析为:
5 - (4 + 1) -- i.e.,(Minus (Num 5) (Plus (Num 4) (Num 1)))
您的评估者将计算出错误的答案0。
但是,如果您真的想使用右关联运算符进行解析,请参见下文。
针对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,ts)
| otherwise = error ("unrecognized token " ++ show t)
parseFactor [] = error "unexpected end of input"
parseTerm :: [String] -> (Ast,ts1) = parseFactor ts
in go f1 ts1
where go acc ("*":ts2)
= let (f2,rest)
parseExpr :: [String] -> (Ast,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,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*