Parsing “x y z” with the precedence of multiply

问题

I'm trying to write a parser for the Mathematica language in F# using FParsec.

I have written one for a MiniML that supports the syntax f x y = (f(x))(y) with high precedence for function application. Now I need to use the same syntax to mean f*x*y and, therefore, have the same precedence as multiply. In particular, x y + 2 = x*y + 2 whereas x y ^ 2 = x * y^2.

How can this be accomplished?

回答1:

As Stephan pointed out in a comment you can split the operator parser into two separate parsers and put your own parser in the middle for space-separated expressions. The following code demonstrates this:

#I "../packages/FParsec.1.0.1/lib/net40-client"
#r "FParsec"
#r "FParsecCS"

open FParsec
open System.Numerics

type Expr =
  | Int of BigInteger
  | Add of Expr * Expr
  | Mul of Expr * Expr
  | Pow of Expr * Expr

let str s = pstring s >>. spaces
let pInt : Parser<_, unit> = many1Satisfy isDigit |>> BigInteger.Parse .>> spaces
let high = OperatorPrecedenceParser<Expr,unit,unit>()
let low = OperatorPrecedenceParser<Expr,unit,unit>()
let pHighExpr = high.ExpressionParser .>> spaces
let pLowExpr = low.ExpressionParser .>> spaces

high.TermParser <-
  choice
    [ pInt |>> Int
      between (str "(") (str ")") pLowExpr ]

low.TermParser <-
  many1 pHighExpr |>> (function [f] -> f | fs -> List.reduce (fun f g -> Mul(f, g)) fs) .>> spaces

low.AddOperator(InfixOperator("+", spaces, 10, Associativity.Left, fun f g -> Add(f, g)))
high.AddOperator(InfixOperator("^", spaces, 20, Associativity.Right, fun f g -> Pow(f, g)))

run (spaces >>. pLowExpr .>> eof) "1 2 + 3 4 ^ 5 6"

The output is:

Add (Mul (Int 1,Int 2),Mul (Mul (Int 3,Pow (Int 4,Int 5)),Int 6))

which represents 1 * 2 + 3 * 4^5 * 6 as expected.

来源：https://stackoverflow.com/questions/29322892/parsing-x-y-z-with-the-precedence-of-multiply

标签

fparsec