Question

几周前我问了一个关于我的ANTLR语法的问题（My simple ANTLR grammar is not working as expected）。自从提出这个问题以来，我已经做了更多的挖掘和调试，并解决了大部分问题。不过，我只留下了一个问题。

我生成的解析器代码没有在处理的文本的一个特定部分中拾取无效标记。词法分析器正确地将事物分解为标记，但在某些情况下，解析器不会触发无效标记。 特别是，当无效令牌位于像“A和”B“这样的短语的末尾时，解析器会忽略它 - 就像令牌甚至不存在一样。

一些具体的例子：

“A和B” - 完全有效
“A @和B” - 解析器正确选择无效的@ token
“A和@B” - 解析器正确选择无效的@ token
“A和B @” - 这就是神秘 - 词法分析器找到@标记，解析器忽略它（！）
“（A和B @）或C” - 更进一步的神秘 - 词法分析器找到@标记，解析器忽略它（！）

这是我的语法：

grammar QvidianPlaybooks; 
options{ language=CSharp3; output=AST; ASTLabelType = CommonTree; } 

public parse
  :  expression
  ;

LPAREN : '(' ;
RPAREN : ')' ;
ANDOR : 'AND'|'and'|'OR'|'or';
NAME : ('A'..'Z');
WS : ' ' { $channel = Hidden; };
THEREST : .;

// ***************** parser rules:
expression : anexpression EOF!;
anexpression : atom (ANDOR^ atom)*;
atom : NAME | LPAREN! anexpression RPAREN!;

然后处理结果树的代码如下所示：

... from the main program
QvidianPlaybooksLexer lexer = new QvidianPlaybooksLexer(new ANTLRStringStream(src));
QvidianPlaybooksParser parser = new QvidianPlaybooksParser(new CommonTokenStream(lexer));
parser.TreeAdaptor = new CommonTreeAdaptor();

CommonTree tree = (CommonTree)parser.parse().Tree;
ValidateTree(tree, 0, iValidIdentifierCount);

// recursive code that walks the tree
public static RuleLogicValidationResult ValidateTree(ITree Tree, int depth, int conditionCount)
{
    RuleLogicValidationResult rlvr = null;

    if (Tree != null)
    {
        CommonErrorNode commonErrorNode = Tree as CommonErrorNode;
        if (null != commonErrorNode)
        {
            rlvr = new RuleLogicValidationResult();
            rlvr.IsValid = false;
            rlvr.ErrorType = LogicValidationErrorType.Other;
            Console.WriteLine(rlvr.ToString());
        }
        else
        {
            string strTree = Tree.ToString();
            strTree = strTree.Trim();
            strTree = strTree.ToUpper();
            if ((Tree.ChildCount != 0) && (Tree.ChildCount != 2))
            {
                rlvr = new RuleLogicValidationResult();
                rlvr.IsValid = false;
                rlvr.ErrorType = LogicValidationErrorType.Other;
                rlvr.InvalidIdentifier = strTree;
                rlvr.ErrorPosition = 0;
                Console.WriteLine(String.Format("CHILD COUNT of {0} = {1}", strTree, tree.ChildCount));
            }
            // if the current node is valid, then validate the two child nodes
            if (null == rlvr || rlvr.IsValid)
            {
                // output the tree node
                for (int i = 0; i < depth; i++)
                {
                    Console.Write("  ");
                }
                Console.WriteLine(Tree);
                rlvr = ValidateTree(Tree.GetChild(0), depth + 1, conditionCount);
                if (rlvr.IsValid)
                {
                    rlvr = ValidateTree(Tree.GetChild(1), depth + 1, conditionCount);
                }
            }
            else
            {
                Console.WriteLine(rlvr.ToString());
            }
        }
    }
    else
    {
        // this tree is null, return a "it's valid" result
        rlvr = new RuleLogicValidationResult();
        rlvr.ErrorType = LogicValidationErrorType.None;
        rlvr.IsValid = true;
    }

    return rlvr;
}

Answer 1

将EOF添加到开始规则的末尾。：）

我的简单ANTLR语法在解析时忽略了某些无效标记

1 个答案: