Question

我正在尝试创建一个简单的反讽解析器。语法应该是

能够识别其中一个或多个

setlink（ “串”）;

它似乎成功识别了1个单一的keyword（）调用，但是连续两个关键字爆炸了。我收到错误状态。当我检查main中的parseTree变量时，我可以看到它在停止解析之前识别的最后一个标记是字母“c”，即“connect”标记中的第一个字母。谁能告诉我这里缺少什么？

code.txt的内容

 setlink("stuffs");
 connect("things");

LilGrammar.cs

 using System;
 using System.Collections.Generic;
 using System.Linq;
 using System.Security;
 using System.Text;
 using System.Threading.Tasks;
 using Irony.Parsing;

 namespace IronyTest
 {
 public class LilGrammar : Grammar
 {
   public LilGrammar()
   {

        //Define terminals

       var program = new NonTerminal("program");
       var command = new NonTerminal("command");
       var commandList = new NonTerminal("commandList");
       var keyword = new NonTerminal("keyword");
       var stringLiteral = new StringLiteral("string", "\"", StringOptions.None);

       var LPAREN = ToTerm("(");
       var RPAREN = ToTerm(")");
       var SEMICO = ToTerm(";");
       var NL = ToTerm("\n");         
       this.Root = program;

       program.Rule = command + Eof;

       keyword.Rule = ToTerm("setlink") | "connect";
       command.Rule = keyword + LPAREN + stringLiteral + RPAREN + SEMICO;  
       commandList.Rule = MakePlusRule(commandList,null , command);
       this.Root = program;

       MarkPunctuation(";");
   }
}
 }

Main.cs

 using System;
 using System.Collections.Generic;
 using System.Linq;
 using System.Text;
 using System.Threading.Tasks;
 using Irony.Parsing;
 using Irony;
 using System.IO;

 namespace IronyTest
 {
 private static void Main(string[] args)
    {
        Grammar grammar = new LilGrammar();
        var parser = new Parser(grammar);
        string path = @"c:\code\code.txt";


        // Open the file to read from. 
        string readText = File.ReadAllText(path);


        //Returns the root to a parase tree
        //var tree = parser.Parse(readText, path);
        ParseTree parseTree = parser.Parse(readText,path);
        ParseTreeNode root = parseTree.Root;
 }

Answer 1

我认为解决方法是让程序包含一个命令列表，而不仅仅是一个命令。

因此，通过以下修改，它似乎正确解析。请注意，我ToTerm("connect")看起来就像keyword.Rule中的setlink一样，我将program.Rule的定义移到了commandList.Rule的定义之后，我改变了program.Rule }引用commandList，而不是command。

keyword.Rule = ToTerm("setlink") | ToTerm("connect");
command.Rule = keyword + LPAREN + stringLiteral + RPAREN + SEMICO;
commandList.Rule = MakePlusRule(commandList, command);
program.Rule = commandList + Eof;
this.Root = program;

Simple Irony Parser无法识别特定匹配项目的多个实例

1 个答案: