我正在编写Haskell中的优先爬升算法,但是由于我不知道的原因,它不起作用。我认为Parsec状态信息在某个时候丢失了,但我甚至不知道这是错误的根源:
module PrecedenceClimbing where
import Text.Parsec
import Text.Parsec.Char
{-
Algorithm
compute_expr(min_prec):
result = compute_atom()
while cur token is a binary operator with precedence >= min_prec:
prec, assoc = precedence and associativity of current token
if assoc is left:
next_min_prec = prec + 1
else:
next_min_prec = prec
rhs = compute_expr(next_min_prec)
result = compute operator(result, rhs)
return result
-}
type Precedence = Int
data Associativity = LeftAssoc
| RightAssoc
deriving (Eq, Show)
data OperatorInfo = OPInfo Precedence Associativity (Int -> Int -> Int)
mkOperator :: Char -> OperatorInfo
mkOperator = \c -> case c of
'+' -> OPInfo 1 LeftAssoc (+)
'-' -> OPInfo 1 LeftAssoc (-)
'*' -> OPInfo 2 LeftAssoc (*)
'/' -> OPInfo 2 LeftAssoc div
'^' -> OPInfo 3 RightAssoc (^)
getPrecedence :: OperatorInfo -> Precedence
getPrecedence (OPInfo prec _ _) = prec
getAssoc :: OperatorInfo -> Associativity
getAssoc (OPInfo _ assoc _) = assoc
getFun :: OperatorInfo -> (Int -> Int -> Int)
getFun (OPInfo _ _ fun) = fun
number :: Parsec String () Int
number = do
spaces
fmap read $ many1 digit
operator :: Parsec String () OperatorInfo
operator = do
spaces
fmap mkOperator $ oneOf "+-*/^"
computeAtom = do
spaces
number
loop minPrec res = (do
oper <- operator
let prec = getPrecedence oper
if prec >= minPrec
then do
let assoc = getAssoc oper
next_min_prec = if assoc == LeftAssoc
then prec + 1
else prec
rhs <- computeExpr(next_min_prec)
loop minPrec $ getFun oper res rhs
else return res) <|> (return res)
computeExpr :: Int -> Parsec String () Int
computeExpr minPrec = (do
result <- computeAtom
loop minPrec result) <|> (computeAtom)
getResult minPrec = parse (computeExpr minPrec) ""由于某种原因,我的程序只是根据情况处理第一个操作或第一个操作数,但没有进一步处理。
GHCi会话:
*PrecedenceClimbing> getResult 1 "46+10"
Right 56
*PrecedenceClimbing> getResult 1 "46+10+1"
Right 56发布于 2016-06-07 15:47:30
我不清楚您的代码到底出了什么问题,但我将提供以下评论:
(1)这些陈述不等于:
Generic Imperative: rhs = compute_expr(next_min_prec)
Haskell: rhs <- computeExpr(next_min_prec)对compute_expr的命令式调用将始终返回。Haskell呼叫可能失败,在这种情况下,调用后的内容永远不会发生。
(2)您实际上是在利用Parsec的优势,一次一次地解析令牌。要了解具有各种先例和关联的运算符的一般解析表达式的"Parsec方式“,请查看:
buildExpression更新
我已经向http://lpaste.net/165651发布了一个解决方案
https://stackoverflow.com/questions/37681474
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