Question

当用户输入与我们定义的规则不匹配时，我需要报告自定义错误。

这是我的代码：

grammar second1;

@lexer::members {
@Override
public void reportError(RecognitionException e) {
    System.out.println("Throwing Exception: "+ e.getMessage());
    throw new IllegalArgumentException(e);
 }
}

@parser::members {
private boolean inbounds(Token t, int min, int max, String methodName) {
   int n = Integer.parseInt(t.getText());    
   if(n >= min && n <= max) {
     return true;
   }
   else {
     System.out.println("The range for value accepted by " + methodName+" is "+min +"-" + max );
     return false;
   }
 }
}

expr       :  SET attribute EOF;

attribute  :  Value1 int1:integer1["Value1"] { System.out.println("Accepted"); }
       |  Value2 integer2 ["Value2"] { System.out.println("Accepted"); }
       ;
exception[int1]: 
        catch[Exception e] {System.out.println("Error Reported for int1");}
exception: 
        catch[Exception e] {System.out.println("General error Reported");}

integer1 [String methodName]   :  Int { inbounds($Int,0,1000,methodName) }? ;
integer2 [String methodName]  :  Int { inbounds($Int,0,10000,methodName) }? ;
Int        :  '0'..'9'+;

SET        :  'set';
Value1     :  'value';
Value2     :  'value2'; 

fragment WS
  : (' ' | '\t')
 ;

但在编译此代码时，我收到以下错误：

error(100): second1.g:26:22: syntax error: antlr: second1.g:26:22: unexpected token: int1  
error(100): second1.g:29:17: syntax error: antlr: second1.g:29:17: unexpected token: :  
error(100): second1.g:32:10: syntax error: antlr: second1.g:32:10: unexpected token: catch  
error(100): second1.g:0:0: syntax error: assign.types: <AST>:0:0: unexpected AST node: <end-of-block>  
error(100): second1.g:0:0: syntax error: assign.types: <AST>:0:0: unexpected end of subtree  
error(100): second1.g:0:0: syntax error: define: <AST>:0:0: unexpected AST node: <end-of-block>  
error(100): second1.g:0:0: syntax error: define: <AST>:0:0: unexpected AST node: <end-of-block>  
error(100): second1.g:0:0: syntax error: define: <AST>:0:0: unexpected end of subtree  
error(106): second1.g:26:27: reference to undefined rule: integer1  
error(106): second1.g:27:22: reference to undefined rule: integer2  
warning(105): second1.g:27:15: no lexer rule corresponding to token: Value2  
warning(105): second1.g:26:15: no lexer rule corresponding to token: Value1  
warning(105): second1.g:24:15: no lexer rule corresponding to token: SET

我该怎么办？ :(
我在网上查了一下，这就是我们在ANTLR 3.x中处理异常的方法为什么它在我的情况下不起作用:(
请帮帮我。

Answer 1

“Catch块不再以'exception'关键字为前缀”，因此您的属性规则为：

attribute  :
      Value1 integer1["Value1"] { System.out.println("Accepted"); }
   |  Value2 integer2["Value2"] { System.out.println("Accepted"); }
   ;
catch[Exception e] {System.out.println("General error Reported");}

接下来，您已经覆盖lexer的reportError方法，而不是解析器之一（调用入站检查）。

要使解析器抛出错误而不是恢复，您可以将reportError复制到@parser :: members部分，然后您可以获得“报告的常规错误”。

但是如果你不想停止antlr的恢复机制，但为了使错误消息更具信息性，你可以从excerpt读取这个免费的The Definitive ANTLR Reference并定义getErrorMessage方法：

public String getErrorMessage(RecognitionException e, String[] tokenNames)
{
    List stack = getRuleInvocationStack(e, this.getClass().getName());
    String msg = null;
    if ( e instanceof NoViableAltException ) {
       NoViableAltException nvae = (NoViableAltException)e;
       msg = " no viable alt; token="+e.token+
          " (decision="+nvae.decisionNumber+
          " state "+nvae.stateNumber+")"+
          " decision=<<"+nvae.grammarDecisionDescription+">>";
    }
    else if(  e instanceof FailedPredicateException  ) {
       FailedPredicateException fpe = (FailedPredicateException)e;
       msg = "failed predicate; token="+fpe.token+
              " (rule="+fpe.ruleName+" predicate="+fpe.predicateText+")";
    }
    else {
       msg = super.getErrorMessage(e, tokenNames);
    }
    return stack+" "+msg;
}
public String getTokenErrorDisplay(Token t) {
    return t.toString();
}

antlr 3.0中的错误处理

1 个答案: