Java的ANTLR 4监听器类的方法不会被调用(Java ANTLR 4 Listener cla

我在ANTLR4做了一个简单的语法解析器与词法分析器和使用Java对象。然后我做了一个监听器子类，并与我的自定义监听器调用ParseTreeWalker。它看起来像监听对象的只有基本方法 - enterEveryRule，exitEveryRule，visitTerminalNode，visitErrorNode - 越来越调用。我的语法有其他规则，其进入/退出功能显然没有得到调用。我在做什么错在这里？这里是我的代码：

import java.lang.StringBuffer;
import java.util.Map;
import java.util.HashMap;
import org.antlr.v4.runtime.*;
import org.antlr.v4.runtime.tree.*;
import java.nio.file.Files;
import java.nio.file.Paths;
import java.util.List;

public class GameLangCompiler {
    public static String Compile(String fileName) {
        ANTLRInputStream input = null;
        try {
            input = new ANTLRFileStream(fileName);
        } catch (Exception ex) {
        ex.printStackTrace();
    }

        GameLangLexer lexer = new GameLangLexer(input);
        CommonTokenStream tokenStream = new CommonTokenStream(lexer);
        GameLangParser parser = new GameLangParser(tokenStream);
        AwesomeListener listener = new AwesomeListener(parser);


        GameLangParser.ProgramContext tree = parser.program();
        ParseTreeWalker.DEFAULT.walk(listener, tree);
        return listener.getSourceCode();
    }

    public static void main(String[] args) {
        if (args.length < 1) {
            System.err.println("Give a filename\n");
            System.exit(1);
        }

        String out = Compile(args[0]);
        System.out.println("-------- Compiler output --------\n" + out);
    }

    /**
     * Code generating listener for GameLang
     */
    public static class AwesomeListener extends GameLangBaseListener {
        private GameLangParser parser;
        private StringBuffer source;
        private Map<String, String> scriptBlocks;
        private static final String UPDATE = "update", RENDER = "render", SETUP = "setup";

        public AwesomeListener(GameLangParser parser) {
            this.parser = parser;
            this.source = new StringBuffer();
            this.scriptBlocks = new HashMap<String, String>();
        }

        public String getSourceCode() {
            return source.toString();
        }

        @Override public void enterProgram(GameLang.ProgramContext ctx) {
            source.append("program\n");
        }

        @Override public void exitProgram(GameLang.ProgramContext ctx) {
            source.append("Entering program!\n");
        }

        @Override public void enterBlock(GameLang.BlockContext ctx) {
            source.append("block\n");
        }

        @Override public void exitBlock(GameLang.BlockContext ctx) { }

        @Override public void enterEveryRule(ParserRuleContext ctx) {
            source.append("Entering a rule:\n");
            source.append(ctx.getText() + "\n\n");
        }
    }
}

那就是越来越解析树走去，编译器的输出：

(program (block setup ( ) (body {\n  x = 5\n})) <EOF>)

-------- Compiler output --------

Entering a rule:
setup(){
  x = 5
}<EOF>

Entering a rule:
setup(){
  x = 5
}

Entering a rule:
{
  x = 5
}

正如你所看到的，只有“enterEveryRule”监听方法获取调用。