Question

我想从我的解析器中获取所有令牌，然后我想过滤输出，得到一个AST列表（myAST）：

    ANTLRStringStream stream = new ANTLRStringStream("P + 1 + F(A + 3)");
    MyLexer lexer = new MyLexer (stream);
    MyParser parser = new MyParser(new CommonTokenStream(lexer));
    LanguajeTreeAdaptor treeAdaptor = new LanguajeTreeAdaptor();
    parser.setTreeAdaptor(treeAdaptor);

我有一个TreeAdaptor用于我的语言：

public class LanguajeTreeAdaptor extends CommonTreeAdaptor{
       public LanguajeTreeAdaptor(){
         super();
       }

  @Override
  public Object create(Token payload) {
      if(payload == null)
         return super.create(payload);

      switch(payload.getType()){
          case EtesGrammarParser.ID:
             return new IdAST(payload);
       ..........
      }
      return super.create(payload);
 }

以下是我语法的一些规则：

program
:   expression EOF!
;

expression
:   exprFinal   
;

exprFinal:  exprFuncCall  |  ID  |  INT  |  DOUBLE  |  STRING  |  exprParenthesis;

exprFuncCall  :  ID LPARENT exprList RPARENT -> ^(FUNC_CALL ID exprList);

所以P，F和A在我的规则中是Id，我正在尝试使用此代码：

    TokenStream input = parser.getTokenStream();
    TokenSource tSource = input.getTokenSource();
    Token currentToken = tSource.nextToken();

    while(currentToken != null){
        System.out.println(currentToken.getText());
        currentToken = tSource.nextToken();
    }

但我想只处理IdAST（我的想法），

    while(currentToken != null){
        if(currentToken instanceof IdAST){
            //call to another method to process the Id
        }
        System.out.println(currentToken.getText());
        currentToken = tSource.nextToken();
    }

我不能这样做因为currentToken是CommonToken，我怎么才能得到那些Id？我正在使用antlr 3 Regards Zinov

如何从antlr中的解析器中获取所有标记

0 个答案: