如何在解析Haskell時正確區分詞幹字符和文字字符

Question

我對Haskell相當新，我試圖做一個簡單的解析器，並使用Parsec模塊。我的解析器的語法是：

data Frag 
    = Lt String 
    | St 
    deriving (Eq, Show) 

type Template = [Frag] 

type FileT = Template 
type CommandT = Template 

data Rule 
    = Rule [FileT] [FileT] [CommandT] 
    deriving (Eq, Show) 

type Makefile = [Rule] 

我已經實現一路下跌至片段（這將是字符集）。下面是我如何試圖處理文本字符的文檔片斷，我不知道如何處理，雖然乾的字符：

template :: Parser [Frag] 
template = 
    do result <- many frag 
     return result 
frag :: Parser Frag 
frag = do Lt x <- (many (noneOf ['\n','\t','\r',':','.'])) 
      return x 

但我發現了這個錯誤，我不知道爲什麼：

Parser\Impl.hs:72:11: error: 
    • Couldn't match expected type ‘[Char]’ with actual type ‘Frag’ 
    • In the pattern: Lt x 
     In a stmt of a 'do' block: 
     Lt x <- (many (noneOf ['\n', '\t', '\r', ':', ....])) 
     In the expression: 
     do { Lt x <- (many (noneOf ['\n', '\t', ....])); 
      return x } 

Parser\Impl.hs:73:11: error: 
    • Couldn't match type ‘[Char]’ with ‘Frag’ 
     Expected type: Text.Parsec.Prim.ParsecT 
         String() Data.Functor.Identity.Identity Frag 
     Actual type: Text.Parsec.Prim.ParsecT 
         String() Data.Functor.Identity.Identity String 
    • In a stmt of a 'do' block: return x 
     In the expression: 
     do { Lt x <- (many (noneOf ['\n', '\t', ....])); 
      return x } 
     In an equation for ‘frag’: 
      frag 
      = do { Lt x <- (many (noneOf ['\n', ....])); 
        return x } 

輸入：

"aaa : bbb ccc" 
"\:aaa : \%bbb \\ccc 

輸出：

[["aaa"] , ["bbb"] , ["ccc"]] 
[[":aaa"] , ["%bbb"] , ["\ccc"]]  

Answer 1

frag :: Parser Frag 
frag = do Lt x <- (many (noneOf ['\n','\t','\r',':','.'])) 
      return x 

many，在這裏，產生了Parser [Char]。您試圖將[Char]結果與Frag模式匹配，導致類型錯誤。相反，你想...

frag :: Parser Frag 
frag = do x <- (many (noneOf ['\n','\t','\r',':','.'])) 
      return (Lt x) 

...或者乾脆：

frag :: Parser Frag 
frag = fmap Lt (many (noneOf ['\n','\t','\r',':','.'])) 

PS：在你的其他定義...

template :: Parser [Frag] 
template = 
    do result <- many frag 
     return result 

...只爲結合result立即使用return就是多餘的。你可以這樣寫：

template :: Parser [Frag] 
template = many frag 

P.P.S：正如你已經注意到了（也許如你預期），many frag是不夠的，做你想要什麼。您將需要以某種方式指定如何劃分您的分段。