f# - 在 FParsec 中解析数字

Question

我已经开始学习 FParsec。它有一种非常灵活的数字解析方式；我可以提供一组我想使用的数字格式:

type Number =
    | Numeral of int
    | Decimal of float
    | Hexadecimal of int
    | Binary of int

let numberFormat = NumberLiteralOptions.AllowFraction
                   ||| NumberLiteralOptions.AllowHexadecimal
                   ||| NumberLiteralOptions.AllowBinary

let pnumber = 
    numberLiteral numberFormat "number"
    |>> fun num -> if num.IsHexadecimal then Hexadecimal (int num.String)
                   elif num.IsBinary then Binary (int num.String)
                   elif num.IsInteger then Numeral (int num.String)
                   else Decimal (float num.String)

但是，我试图解析的语言有点奇怪。数字可以是数字(非负 int)、十进制(非负 float)、十六进制(带前缀 #x)或二进制(带前缀 #b):

numeral: 0, 2
decimal: 0.2, 2.0
hexadecimal: #xA04, #x611ff
binary: #b100, #b001

现在我必须通过替换 # 进行两次解析通过 0 (如有必要)使用 pnumber :

let number: Parser<_, unit> =  
    let isDotOrDigit c = isDigit c || c = '.'
    let numOrDec = many1Satisfy2 isDigit isDotOrDigit 
    let hexOrBin = skipChar '#' >>. manyChars (letter <|> digit) |>> sprintf "0%s"
    let str = spaces >>. numOrDec <|> hexOrBin
    str |>> fun s -> match run pnumber s with
                     | Success(result, _, _)   -> result
                     | Failure(errorMsg, _, _) -> failwith errorMsg

在这种情况下，有什么更好的解析方法？或者我怎样才能改变 FParsec 的 CharStream能够使条件解析更容易？

Answer 1

如果您想生成良好的错误消息并正确检查溢出，解析数字可能会非常困惑。

以下是您的数字解析器的简单 FParsec 实现:

let numeralOrDecimal : Parser<_, unit> =
    // note: doesn't parse a float exponent suffix
    numberLiteral NumberLiteralOptions.AllowFraction "number" 
    |>> fun num -> 
            // raises an exception on overflow
            if num.IsInteger then Numeral(int num.String)
            else Decimal(float num.String)

let hexNumber =    
    pstring "#x" >>. many1SatisfyL isHex "hex digit"
    |>> fun hexStr -> 
            // raises an exception on overflow
            Hexadecimal(System.Convert.ToInt32(hexStr, 16)) 

let binaryNumber =    
    pstring "#b" >>. many1SatisfyL (fun c -> c = '0' || c = '1') "binary digit"
    |>> fun hexStr -> 
            // raises an exception on overflow
            Binary(System.Convert.ToInt32(hexStr, 2))


let number =
    choiceL [numeralOrDecimal
             hexNumber
             binaryNumber]
            "number literal"

在溢出时生成好的错误消息会使这个实现有点复杂，因为理想情况下你还需要在错误之后回溯，以便错误位置最终出现在数字文字的开头(参见 numberLiteral 文档以获取示例) .

优雅地处理可能的溢出异常的一个简单方法是使用一个小的异常处理组合器，如下所示:

let mayThrow (p: Parser<'t,'u>) : Parser<'t,'u> =
    fun stream ->
        let state = stream.State        
        try 
            p stream
        with e -> // catching all exceptions is somewhat dangerous
            stream.BacktrackTo(state)
            Reply(FatalError, messageError e.Message)

然后你可以写

let number = mayThrow (choiceL [...] "number literal")

我不确定您所说的“更改 FParsec 的 CharStream 以使条件解析更容易”是什么意思，但以下示例演示了如何编写仅使用 CharStream 的低级实现方法直接。

type NumberStyles = System.Globalization.NumberStyles
let invariantCulture = System.Globalization.CultureInfo.InvariantCulture

let number: Parser<Number, unit> =
  let expectedNumber = expected "number"
  let inline isBinary c = c = '0' || c = '1'
  let inline hex2int c = (int c &&& 15) + (int c >>> 6)*9

  let hexStringToInt (str: string) = // does no argument or overflow checking        
      let mutable n = 0
      for c in str do
          n <- n*16 + hex2int c
      n    

  let binStringToInt (str: string) = // does no argument or overflow checking
      let mutable n = 0
      for c in str do
          n <- n*2 + (int c - int '0')
      n

  let findIndexOfFirstNonNull (str: string) =
      let mutable i = 0
      while i < str.Length && str.[i] = '0' do
          i <- i + 1
      i

  let isHexFun = id isHex // tricks the compiler into caching the function object
  let isDigitFun = id isDigit
  let isBinaryFun = id isBinary

  fun stream ->
    let start = stream.IndexToken
    let cs = stream.Peek2()        
    match cs.Char0, cs.Char1 with
    | '#', 'x' ->
        stream.Skip(2)
        let str = stream.ReadCharsOrNewlinesWhile(isHexFun, false)
        if str.Length <> 0 then
            let i = findIndexOfFirstNonNull str
            let length = str.Length - i
            if length < 8 || (length = 8 && str.[i] <= '7') then
                Reply(Hexadecimal(hexStringToInt str))
            else
                stream.Seek(start)
                Reply(Error, messageError "hex number literal is too large for 32-bit int")
        else 
            Reply(Error, expected "hex digit")

    | '#', 'b' ->
        stream.Skip(2)
        let str = stream.ReadCharsOrNewlinesWhile(isBinaryFun, false)
        if str.Length <> 0 then
            let i = findIndexOfFirstNonNull str
            let length = str.Length - i
            if length < 32 then 
                Reply(Binary(binStringToInt str))
            else
                stream.Seek(start)
                Reply(Error, messageError "binary number literal is too large for 32-bit int")
        else 
            Reply(Error, expected "binary digit")

    | c, _ ->
        if not (isDigit c) then Reply(Error, expectedNumber)
        else
            stream.SkipCharsOrNewlinesWhile(isDigitFun) |> ignore
            if stream.Skip('.') then
                let n2 = stream.SkipCharsOrNewlinesWhile(isDigitFun)
                if n2 <> 0 then
                    // we don't parse any exponent, as in the other example
                    let mutable result = 0.
                    if System.Double.TryParse(stream.ReadFrom(start), 
                                              NumberStyles.AllowDecimalPoint,
                                              invariantCulture, 
                                              &result)
                    then Reply(Decimal(result))
                    else 
                        stream.Seek(start)
                        Reply(Error, messageError "decimal literal is larger than System.Double.MaxValue")                    
                else 
                    Reply(Error, expected "digit")
            else
               let decimalString = stream.ReadFrom(start)
               let mutable result = 0
               if System.Int32.TryParse(stream.ReadFrom(start),
                                        NumberStyles.None,
                                        invariantCulture,
                                        &result)
               then Reply(Numeral(result))
               else 
                   stream.Seek(start)
                   Reply(Error, messageError "decimal number literal is too large for 32-bit int")

虽然此实现在没有系统方法帮助的情况下解析十六进制和二进制数，但它最终将十进制数的解析委托(delegate)给 Int32.TryParse 和 Double.TryParse 方法。

正如我所说:这很困惑。