使用解析器组合器时堆栈溢出 [英] Stack overflow when using parser combinators

问题描述

import scala.util.parsing.combinator._

object ExprParser extends JavaTokenParsers {
    lazy val name: Parser[_] = "a" ~ rep("a" | "1") | function_call
    lazy val function_call = name ~ "(" ~> name <~ ")"
}

为 function_call.parseAll("aaa(1)") 无限重复.很明显，是因为1不能插入name，name进入function_call，它尝试name，进入funciton调用.您如何解决此类情况?

recurs indefinitely for function_call.parseAll("aaa(1)"). Obviously, it is because 1 cannot inter the name and name enters the function_call, which tries the name, which enters the funciton call. How do you resolve such situations?

有一个将名称简化为简单标识符的解决方案

There was a solution to reduce name to simple identifier

def name = rep1("a" | "1") 
def function_call = name ~ "(" ~ (function_call | name) ~ ")"

但我不想这样做，因为 name ::= identifier |function_call 在 VHDL 规范中是 BNF 化的，function_call 可能在其他地方共享.左递归消除在这里找到是出于同样的原因不受欢迎

but I prefer not to do this because name ::= identifier | function_call is BNF-ed in VHDL specification and function_call is probably shared elsewhere. The left recursion elimination found here is undesirable for the same reason

def name: Parser[_] = "a" ~ rep("a" | "1") ~ pared_name
def pared_name: Parser[_] = "(" ~> name <~ ")" | ""

顺便说一句，我也想知道，如果我修复了这个错误，name.parseAll 是只将aaa"作为名称规则中的第一个选项还是使用整个aaa(1)"?在只消耗 aaa 之前，如何命名以消耗整个 aaa(1)?我想我应该将 function_call 作为名称中的第一个选项，但在这种情况下它会更急切地堆栈溢出?

BTW, I also wonder, if I fix the error, will name.parseAll consume "aaa" only as first alternative in the name rule or take whole "aaa(1)"? How can I make name to consume the whole aaa(1) before consuming only aaa? I guess that I should put function_call a first alternative in the name but it will stack overflow even more eagerly in this case?

推荐答案

一个简单的解决方案是使用 Packrat 解析器:

An easy solution is use the packrat parser:

object ExprParser extends JavaTokenParsers with PackratParsers {
  lazy val name: PackratParser[_] = "a" ~ rep("a" | "1") | function_call
  lazy val function_call: PackratParser[_] = name ~ "(" ~> name <~ ")"
}

输出:

scala> ExprParser.parseAll(ExprParser.function_call, "aaa(1)")
res0: ExprParser.ParseResult[Any] =
[1.5] failure: Base Failure

aaa(1)
    ^

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