Scala 解析器组合器：如何解析“if(x)”如果 x 可以包含“)”

发布于 2024-09-02 14:29:14 字数 978 浏览 3 评论 0原文

我正在尝试让它工作：

def emptyCond: Parser[Cond] = ("if" ~ "(") ~> regularStr <~ ")" ^^ { case s => Cond("",Nil,Nil) }

其中 regularStr 被定义为接受许多内容，包括“)”。当然，我希望这是一个可接受的输入：if(foo())。但对于任何 if(x) ，它都会将“)”作为 regularStr 的一部分，因此这个解析器永远不会成功。

我缺少什么？

编辑：

regularStr 不是正则表达式。它是这样定义的：

  def regularStr = rep(ident | numericLit | decimalLit | stringLit | stmtSymbol) ^^ { case s => s.mkString(" ") }

符号是：

  val stmtSymbol = "*" | "&" | "." | "::" | "(" | ")" | "*" | ">=" | "<=" | "=" | 
               "<" | ">" | "|" | "-" | "," | "^" | "[" | "]" | "?" | ":" | "+" |
               "-=" | "+=" | "*=" | "/=" | "&&" | "||" | "&=" | "|="

我不需要详尽的语言检查，只需要控制结构。所以我并不关心 if() 中的“()”里面有什么，我想接受任何标识符、符号等序列。因此，就我的目的而言，即使 if())) 也应该有效，其中“))”是 if 的“条件”。

原文

I'm trying to get this to work:

def emptyCond: Parser[Cond] = ("if" ~ "(") ~> regularStr <~ ")" ^^ { case s => Cond("",Nil,Nil) }

where regularStr is defined to accept a number of things, including ")". Of course, I want this to be an acceptable input: if(foo()). But for any if(x) it is taking the ")" as part of the regularStr and so this parser never succeeds.

What am I missing?

Edit:

regularStr is not a regular expression. It is defined thus:

  def regularStr = rep(ident | numericLit | decimalLit | stringLit | stmtSymbol) ^^ { case s => s.mkString(" ") }

and the symbols are:

  val stmtSymbol = "*" | "&" | "." | "::" | "(" | ")" | "*" | ">=" | "<=" | "=" | 
               "<" | ">" | "|" | "-" | "," | "^" | "[" | "]" | "?" | ":" | "+" |
               "-=" | "+=" | "*=" | "/=" | "&&" | "||" | "&=" | "|="

I don't need exhaustive language check, just the control structures. So I don't really care what's inside "()" in if(), I want to accept any sequence of identifiers, symbols, etc. So, for my purposes even if())) should be valid, where "))" is the if's "condition".

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故事和酒 2024-09-09 14:29:14

正则表达式无法识别具有嵌套、平衡结构的语言，例如 (...)、[...]、{...}< /code> 等。因此，您将需要使用进一步的上下文无关产生式（不是正则表达式）来匹配 regularStr 部分。

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烟沫凡尘 2024-09-09 14:29:14

好吧，接受 if())) 并不是真正的要求，只是我愿意接受的一个例子，以便使我的解析尽可能便宜，只需要担心捕获控制结构。

但看来我不能这么便宜而且仍然可以工作。因此，由于 if() 构造有括号，所以我所要做的就是期望里面的括号具有良好平衡的括号。不期望的结尾“)”不能成为条件的一部分。

我这样做了：

  val regularNoParens = ident | numericLit | decimalLit | stringLit | stmtSymbol 
  def regularParens: Parser[String] = "(" ~ rep(regularNoParens | regularParens) ~ ")" ^^ { case l ~ s ~ r => l + s.mkString(" ") + r } 
  def regularStr = rep(regularNoParens | regularParens) ^^ { case s => s.mkString(" ") }

我从 stmtSymbol 中取出了“(”和“)”。作品！

编辑：它不支持嵌套，已修复。

OK, accepting if())) was not really a requirement, just an example of what I would be willing to accept in order to make my parsing as cheap as possible, to just worry about capturing control structures.

However it appears I can't be so cheap and still have it work. So, since the if() construct has parenthesis, all I have to do is expect what's inside to have well balanced parenthesis. A closing ")" where one isn't expected cannot be part of the condition.

I did this:

  val regularNoParens = ident | numericLit | decimalLit | stringLit | stmtSymbol 
  def regularParens: Parser[String] = "(" ~ rep(regularNoParens | regularParens) ~ ")" ^^ { case l ~ s ~ r => l + s.mkString(" ") + r } 
  def regularStr = rep(regularNoParens | regularParens) ^^ { case s => s.mkString(" ") }

And I took out "(" and ")" from stmtSymbol. Works!

Edit: it didn't support nesting, fixed it.

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