Question

我希望下面的代码返回[LoadInt 1,LoadDub 2.5,LoadInt 3]，但是在解析[LoadInt 1,LoadDub 2]并面向.5,3之后，它会失败。我该如何做才能使它必须一直解析到逗号才能成功进行解析，并且对int进行2.5解析会失败？

import qualified Data.Attoparsec.ByteString.Char8 as A
import Data.Attoparsec.ByteString.Char8 (Parser)
import Data.ByteString.Char8 (pack)
import Data.Attoparsec.Combinator
import Control.Applicative ((*>),(<$>),(<|>))
data LoadNum = LoadInt Int | LoadDub Double deriving (Show)

someFunc :: IO ()
someFunc = putStrLn . show $ A.parseOnly (lnParser <* A.endOfInput) (pack testString)


testString :: String
testString = "1,2.5,3"

lnParser :: Parser [LoadNum]
lnParser = (sepBy1' (ld <* A.atEnd) (A.char ','))

double :: Parser Double
double = A.double

int :: Parser Int
int = A.signed A.decimal

ld :: Parser LoadNum
ld = ((LoadInt <$> int ) <|> (LoadDub <$> double))

Answer 1

您可以使用一小段前瞻性来确定是否到达列表元素的末尾。所以：

int :: Parser Int
int = do
    i <- A.signed A.decimal
    next <- A.peekChar
    case next of
        Nothing -> pure i
        Just ',' -> pure i
        _ -> fail "nah"

Attoparsec-确保sepBy1消耗了全部内容

1 个答案: