ANTLR java测试文件无法创建树语法对象 [英] ANTLR java test file can't create object of tree grammar

问题描述

我正在使用针对java的ANTLR 3.x创建一个解析器。我编写了解析器语法（用于创建抽象语法树，AST）和树语法（用于在AST上执行操作）。最后，为了测试两个语法文件，我用Java编写了一个测试文件。

I am creating a parser using ANTLR 3.x that targets java. I have written both parser grammar (for creating Abstract Syntax Tree, AST) and Tree Grammar (for performing operations on AST). Finally, to test both grammar files, I have written a test file in Java.

看看下面的代码，

协议语法

grammar protocol;
options {
      language = Java;
  output = AST;
}

tokens{ //imaginary tokens
PROT;
INITIALP;
PROC;
TRANSITIONS;
}
@header {
import twoprocess.Configuration;
package com.javadude.antlr3.x.tutorial;
}

@lexer::header {
  package com.javadude.antlr3.x.tutorial;
}
/*
parser rules, in lowercase letters
*/
program
    : declaration+
    ;
declaration
    :protocol
    |initialprocess
    |process
    |transitions
    ;

protocol
    :'protocol' ID ';' -> ^(PROT ID)
    ;
initialprocess
    :'pin' '=' INT ';' -> ^(INITIALP INT)
    ;
process
    :'p' '=' INT ';' -> ^(PROC INT)
    ;
transitions
    :'transitions' '=' INT ('(' INT ',' INT ')') + ';' -> ^(TRANSITIONS INT INT INT*)
    ;

/*
lexer rules (tokens), in upper case letters
*/
ID  
    : (('a'..'z' | 'A'..'Z'|'_')('a'..'z' | 'A'..'Z'|'0'..'9'|'_'))*;
INT 
    : ('0'..'9')+;
WHITESPACE
    : ('\t' | ' ' | '\r' | '\n' | '\u000C')+ {$channel = HIDDEN;};

protocolWalker

grammar protocolWalker;

options {
  language = Java;
  //Error, eclipse can't access tokenVocab named protocol
  tokenVocab = protocol;    //import tokens from protocol.g i.e, from protocol.tokens file
  ASTLabelType = CommonTree;
  }

@header {
import twoprocess.Configuration;
package com.javadude.antlr3.x.tutorial;
}

program
    : declaration+
    ;

declaration
    :protocol
    |initialprocess
    |process
    |transitions
    ;

protocol
    :^(PROT ID)
    {System.out.println("create protocol " +$ID.text);}
    ;

initialprocess
    :^(INITIALP INT)
    {System.out.println("");}
    ;

process
    :^(PROC INT)
    {System.out.println("");}
    ;

transitions
    :^(TRANSITIONS INT INT INT*)
    {System.out.println("");}
    ;

Protocoltest.java

package com.javadude.antlr3.x.tutorial;  
import org.antlr.runtime.*;  
import org.antlr.runtime.tree.*;  
import org.antlr.runtime.tree.CommonTree;  
import org.antlr.runtime.tree.CommonTreeNodeStream;  
public class Protocoltest { 



/**
 * @param args
 */
public static void main(String[] args) throws Exception {
    //create input stream from standard input
    ANTLRInputStream input = new ANTLRInputStream(System.in);
    //create a lexer attached to that input stream
    protocolLexer lexer = new protocolLexer(input);
    //create a stream of tokens pulled from the lexer
    CommonTokenStream tokens = new CommonTokenStream(lexer);

    //create a pareser attached to teh token stream
    protocolParser parser = new protocolParser(tokens);
    //invoke the program rule in get return value
    protocolParser.program_return r =parser.program();
    CommonTree t = (CommonTree)r.getTree();
    //output the extracted tree to the console
    System.out.println(t.toStringTree());

    //walk resulting tree; create treenode stream first
    CommonTreeNodeStream nodes = new CommonTreeNodeStream(t);
    //AST nodes have payloads that point into token stream
    nodes.setTokenStream(tokens);


    //create a tree walker attached to the nodes stream  
            //Error, can't create TreeGrammar object called walker
    protocolWalker walker = new protocolWalker(nodes);

    //invoke the start symbol, rule program
    walker.program();
    }
}

问题：

1. 在protocolWalker中，我无法访问令牌（protocol.tokens）

1. In protocolWalker, I can't access the tokens (protocol.tokens)

//Error, eclipse can't access tokenVocab named protocol  
    tokenVocab = protocol; //import tokens from protocol.g i.e, from protocol.tokens file

In in protocolWalker ，我可以在动作列表中创建名为Configuration的java类对象吗？

In In protocolWalker, can I create the object of java class, called Configuration, in the action list?

protocol
    :^(PROT ID)
       {System.out.println("create protocol " +$ID.text);
        Configuration conf = new Configuration();
       }
    ;

在Protocoltest.java中

In Protocoltest.java

//create a tree walker attached to the nodes stream    
//Error, can't create TreeGrammar object called walker  
    protocolWalker walker = new protocolWalker(nodes);  

无法创建protocolWalker对象。我在示例和教程中已经看到创建了这样的对象。

Object of protocolWalker can't be created. I have seen in the examples and the tutorials that such object is created.

推荐答案

在protocolWalker中，我可以' t访问令牌（protocol.tokens）...

In protocolWalker, I can't access the tokens (protocol.tokens)...

它似乎正在访问 protocol.tokens 罚款：将 tokenVocab 更改为其他内容会产生一个现在不会产生的错误。 protocolWalker.g的问题在于它被定义为令牌解析器（语法protocolWalker ），但它被用作树解析器。将语法定义为树语法protocolWalker 消除了我看到的关于未定义标记的错误。

It seems to be accessing protocol.tokens fine: changing tokenVocab to something else produces an error that it doesn't produce now. The problem with protocolWalker.g is that it's defined as a token parser (grammar protocolWalker) but it's being used like a tree parser. Defining the grammar as tree grammar protocolWalker took away the errors that I was seeing about the undefined tokens.

在protocolWalker中，我可以在动作列表中创建名为Configuration的java类对象吗？

In protocolWalker, can I create the object of java class, called Configuration, in the action list?

是的，您可以。正常的Java编程警告适用于导入类等等，但它可以像 System.out.println 这样的代码使用。

Yes, you can. The normal Java programming caveats apply about importing the class and so on, but it's as available to you as code like System.out.println.

在Protocoltest.java中...无法创建protocolWalker的对象。

In Protocoltest.java ... Object of protocolWalker can't be created.

protocolWalker.g（现在）生成名为 protocolWalkerParser 。当您将其更改为树语法时，它将生成一个名为 protocolWalker 的树解析器。

protocolWalker.g (as it is now) produces a token parser named protocolWalkerParser. When you change it to a tree grammar, it'll produce a tree parser named protocolWalker instead.

谢谢很多用于发布整个语法。这使得回答问题变得更加容易。

Thanks a lot for posting the whole grammars. That made answering the question much easier.

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