如果从文件读取，秒差距失败且没有错误

我编写了一个小型秒差距解析器来从用户提供的输入字符串或输入文件中读取样本。如果输入以分号分隔的字符串形式提供，则它会在错误输入时正确失败，并显示有用的错误消息：

> readUncalC14String "test1,7444,37;6800,36;testA,testB,2000,222;test3,7750,40"
*** Exception: Error in parsing dates from string: (line 1, column 29):
unexpected "t"
expecting digit

但输入文件却默默失败inputFile.txt具有相同的条目：

test1,7444,37
6800,36
testA,testB,2000,222
test3,7750,40

> readUncalC14FromFile "inputFile.txt"
[UncalC14 "test1" 7444 37,UncalC14 "unknownSampleName" 6800 36]

为什么会这样以及我该如何做readUncalC14FromFile也以一种有用的方式失败？

这是我的代码的最小子集：

import qualified Text.Parsec                    as P
import qualified Text.Parsec.String             as P

data UncalC14 = UncalC14 String Int Int deriving Show

readUncalC14FromFile :: FilePath -> IO [UncalC14]
readUncalC14FromFile uncalFile = do
    s <- readFile uncalFile
    case P.runParser uncalC14SepByNewline () "" s of
        Left err -> error $ "Error in parsing dates from file: " ++ show err
        Right x -> return x
    where
        uncalC14SepByNewline :: P.Parser [UncalC14]
        uncalC14SepByNewline = P.endBy parseOneUncalC14 (P.newline <* P.spaces)

readUncalC14String :: String -> Either String [UncalC14]
readUncalC14String s = 
    case P.runParser uncalC14SepBySemicolon () "" s of
        Left err -> error $ "Error in parsing dates from string: " ++ show err
        Right x -> Right x
    where 
        uncalC14SepBySemicolon :: P.Parser [UncalC14]
        uncalC14SepBySemicolon = P.sepBy parseOneUncalC14 (P.char ';' <* P.spaces)

parseOneUncalC14 :: P.Parser UncalC14
parseOneUncalC14 = do
    P.try long P.<|> short
    where
        long = do
            name <- P.many (P.noneOf ",")
            _ <- P.oneOf ","
            mean <- read <$> P.many1 P.digit
            _ <- P.oneOf ","
            std <- read <$> P.many1 P.digit
            return (UncalC14 name mean std)
        short = do
            mean <- read <$> P.many1 P.digit
            _ <- P.oneOf ","
            std <- read <$> P.many1 P.digit
            return (UncalC14 "unknownSampleName" mean std)

这里发生的情况是您输入的前缀是有效的字符串。要强制秒差距使用整个输入，您可以使用eof parser:

uncalC14SepByNewline = P.endBy parseOneUncalC14 (P.newline <* P.spaces) <* P.eof

一个有效而另一个无效的原因是由于两者之间的差异sepBy and endBy。这是一个更简单的例子：

sepTest, endTest :: String -> Either P.ParseError String
sepTest s = P.runParser (P.sepBy (P.char 'a') (P.char 'b')) () "" s
endTest s = P.runParser (P.endBy (P.char 'a') (P.char 'b')) () "" s

以下是一些有趣的例子：

ghci> sepTest "abababb"
Left (line 1, column 7):
unexpected "b"
expecting "a"

ghci> endTest "abababb"
Right "aaa"

ghci> sepTest "ababaa"
Right "aaa"

ghci> endTest "ababaa"
Left (line 1, column 6):
unexpected "a"
expecting "b"

正如你所看到的sepBy and endBy可能会默默地失败，但是sepBy如果前缀不以分隔符结尾，则静默失败b and endBy如果前缀不在主解析器中结束，则静默失败a.

所以你应该使用eof如果您想确保读取整个文件/字符串，请在两个解析器之后。