2

我正在尝试创建一个简单的 Irony 解析器。语法应该是

能够识别其中一项或多项

设置链接(“字符串”);

它似乎成功识别了 1 个关键字()调用,但连续两个调用失败了。我得到一个错误状态。当我在 main 中检查 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。谁能告诉我我在这里缺少什么?

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(";");
   }
}
 }

主文件

 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;
 }
4

1 回答 1

4

我认为解决方法是使程序由一系列命令组成,而不仅仅是一个命令。

因此,通过以下修改,它似乎可以正确解析。请注意,我ToTerm("connect")keyword.Ruleand 的定义中将 的定义移至 and 的定义program.Rule之后commandList.Rule,我将其更改program.Rule为引用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;
于 2015-05-06T22:18:41.740 回答