Attoparsec

Question

我想将下面的代码 return [LoadInt 1,LoadDub 2.5,LoadInt 3]，但在解析 [LoadInt 1,LoadDub 2] 并面对 .5,3 后它失败了。我该怎么做才能使它必须一直解析到逗号才能成功解析，而 2.5 上的 int 解析失败？

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

您可以使用一点点前瞻来确定您是否到达了列表元素的末尾。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 - 确保使用 sepBy1 消耗全部内容

Attoparsec - ensure entire contents consumed with sepBy1

haskell