对于@BartKiers 建议的问题和语法(谢谢!),我添加了options
块以指定输出为
options{
language=Java;
output=AST;
ASTLabelType=CommonTree;
}
但是,我无法弄清楚如何访问输出,即AST
. 我需要遍历树并处理输入中指定的每个操作。
在这里使用您的示例,我正在尝试实现返回值的规则。但是,我遇到了以下错误:
relational returns [String val]
: STRINGVALUE ((operator)^ term)?
{val = $STRINGVALUE.text + $operator.text + $term.text; }
;
term returns [String rhsOperand]
: QUOTEDSTRINGVALUE {rhsOperand = $QUOTEDSTRINGVALUE.text;}
| NUMBERVALUE {rhsOperand = $NUMBERVALUE.text; }
| '(' condition ')'
;
编译错误:
Checking Grammar RuleGrammarParser.g...
\output\RuleGrammarParser.java:495: cannot find symbol
symbol : variable val
location: class RuleGrammarParser
val = (STRINGVALUE7!=null?STRINGVALUE7.getText():null) + (operator8!=null?input.toString(operator8.start,operator8.stop):null) + (term9!=null?input.toString(term9.start,term9.stop):null);
^
\output\RuleGrammarParser.java:612: cannot find symbol
symbol : variable rhsOperand
location: class RuleGrammarParser
rhsOperand = (QUOTEDSTRINGVALUE10!=null?QUOTEDSTRINGVALUE10.getText():null);
^
\output\RuleGrammarParser.java:632: cannot find symbol
symbol : variable rhsOperand
location: class RuleGrammarParser
rhsOperand = (NUMBERVALUE11!=null?NUMBERVALUE11.getText():null);
^
3 errors
你能帮我理解为什么编译失败吗?
添加了 pastebin:http://pastebin.com/u1Bv3L0A