[英]FParsec combinator to turn Parser<char,_> until Parser<string,_>?
I'm certain there's a really simple answer to this, but I've been staring at this all day and I can't figure it out. 我确信这有一个非常简单的答案,但我一整天都在盯着这个,我无法弄明白。
As per the tutorial, I'm implementing a JSON parser. 根据教程,我正在实现一个JSON解析器。 To challenge myself, I'm implementing the
number
parser myself. 为了挑战自己,我自己实现了
number
解析器。
This is what I got so far: 这是我到目前为止所得到的:
let jnumber =
let neg = stringReturn "-" -1 <|> preturn 1
let digit = satisfy (isDigit)
let digit19 = satisfy (fun c -> isDigit c && c <> '0')
let digits = many1 digit
let ``int`` =
digit
<|> (many1Satisfy2 (fun c -> isDigit c && c <> '0') isDigit)
The trouble is that digit
is a Parser<char,_>
, whereas the second option for int
is a Parser<string,_>
. 问题是
digit
是Parser<char,_>
,而int
的第二个选项是Parser<string,_>
。 Would I normally just use a combinator to turn digit
into a Parser<char,_>
, or is there something else I should do? 我通常只会使用组合器将
digit
转换为Parser<char,_>
,还是我还应该做些什么呢?
The |>>
operator is what you're looking for. |>>
运算符是您正在寻找的。 I quote the FParsec reference : 我引用FParsec参考 :
val (|>>): Parser<'a,'u> -> ('a -> 'b) -> Parser<'b,'u>
The parser p |>> f applies the parser p and returns the result of the function application fx, where x is the result returned by p.
解析器p | >> f应用解析器p并返回函数应用程序fx的结果,其中x是p返回的结果。
p |>> f is an optimized implementation of p >>= fun x -> preturn (fx).
p | >> f是p >> = fun x - > preturn(fx)的优化实现。
For example: 例如:
let jnumber =
let neg = stringReturn "-" -1 <|> preturn 1
let digit = satisfy (isDigit)
let digit19 = satisfy (fun c -> isDigit c && c <> '0')
let digits = many1 digit
(digit |>> string) (* The operator is used here *)
<|> (many1Satisfy2 (fun c -> isDigit c && c <> '0') isDigit)
You may want to read FParsec tutorial on parsing JSON which uses this operator quite intensively. 您可能希望阅读解析JSON的FParsec教程,该教程非常密集地使用此运算符。
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