忽略PEG.js的空白 [英] Ignore whitespace with PEG.js

问题描述

我想用我的语法忽略空白和换行，以便

I want to ignore whitespaces and new lines with my grammar so they are missing in the PEG.js output. Also, a literal within brackets should be returned in a new array.

语法

start
  = 'a'? sep+ ('cat'/'dog') sep* '(' sep* stmt_list sep* ')'

stmt_list
  = exp: [a-zA-Z]+ { return new Array(exp.join('')) }

sep
  = [' '\t\r\n]

测试用例

a dog( Harry )

输出

[
   "a",
   [
      " "
   ],
   "dog",
   [],
   "(",
   [
      " "
   ],
   [
       "Harry"
   ],
   [
      " "
   ],
   ")"
]

我想要的输出

[
   "a",
   "dog",
   [
      "Harry"
   ]
]

推荐答案

您必须使用更多的非终结符"来分解语法(不确定是否在PEG中称呼它们):

You have to break up the grammar more, using more "non-terminals" (not sure if that's what you call them in a PEG):

start
  = article animal stmt_list

article
  = article:'a'? __ { return article; }

animal
  = animal:('cat'/'dog') _ { return animal; }

stmt_list
  = '(' _ exp:[a-zA-Z]+ _ ')' { return [ exp.join('') ]; }

// optional whitespace
_  = [ \t\r\n]*

// mandatory whitespace
__ = [ \t\r\n]+

感谢您提出这个问题！

修改: 要提高可读性，请使用以下两种形式:_和__

To increase readability, have two productions: _ and __

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