java - 在 ANTLR 中分离多个规则时出错

Question

我几乎完成了与 ANTLR 的第一次冒险，这是一次相当不错的旅行。不幸的是，总是只计入马蹄铁、手榴弹和核武器，对吧？

无论如何，我正在尝试解析如下所示的输入:

; IF AGE IS LESS THAN 21, STILL RETURN TRUE FOR OVERSEAS LOCATION \r\n
SHOW "AGE REQUIREMENTS FAILED" FOR \r\n
IF AGE < 21 THEN \r\n
LOCATION = "OVERSEAS" \r\n
ENDIF \r\n
\r\n
; NEED  SOMEONE WHO HAS WORKED FOR US FOR > 1 YEAR EXCEPT FOR CEO \r\n
SHOW "MINIMUM TIME REQUIREMENT NOT MET" FOR \r\n
IF STARTDATE > TODAY - 1 YEAR THEN \r\n
EMPLID=001 \r\n
ENDIF \r\n

通常，如果测试失败，则会显示该消息。

无论如何，一个集合可以包含一个或多个 SHOW 规则。处理单个 SHOW 规则有效，但当输入流包含 > 1 个 SHOW 规则时，它不会“拆分”。

以下是语法中的相关规则:

showGroup returns [List<PolicyEvaluation> value]
  @init {List<PolicyEvaluation> peList = new ArrayList<PolicyEvaluation>();}
  : (expr1=show)* {peList.add($expr1.value);}
  {
    System.out.println("Entered policyGroup rule");
    $value = peList;
  }
  ;

// evaluate a single SHOW statement
show returns [PolicyEvaluation value]
  : ('SHOW' expr1=STRING 'FOR')? expr2=ifStatement EOL*
  {
    System.out.println("Entered show rule");
    Boolean expr2Value = (Boolean) $expr2.value;
    PolicyEvaluation pe = new PolicyEvaluation();
    if (expr1 == null) {
      pe.setValue(expr2Value);
      pe.setMessage(null);
    } else {
        if (expr2Value == false) {
          pe.setValue(false);
          pe.setMessage(expr1.getText());
        } else {
          pe.setValue(true);
          pe.setMessage(null);
        }   
    }
    $value = pe;
  }
  ;
// rules leading up to the show rule
// domain-specific grammar rules
STRING: '"' ID  (' ' ID)* '"'
  {
    System.out.println("Entered STRING lexer rule");
    // strip the quotes once we match this token
    setText(getText().substring(1, getText().length()-1));
  }
  ;
COMMENT: ';' (ID|' ')* EOL {$channel = HIDDEN;};
EOL: ('\r'|'\n'|'\r\n') {$channel = HIDDEN;};
SPACE: ' ' {$channel = HIDDEN;};

也许这很简单。任何帮助表示赞赏。

杰森

Answer 1

尝试改变这个:(expr1=show)* {peList.add($expr1.value);}
对此:(expr1=show {peList.add($expr1.value);})*
原来的 Action 只会触发show匹配已完成，让您操作最后一个 expr1 .