遍历AST树并获得子节点值Irony

本文关键字：子节点 Irony AST 遍历 | 更新日期: 2023-09-27 18:05:18

我想通过AST树并获得ChildNode等于"IfStatement"，但可能有很多ChildNode，如下面的代码。是否有一些方法可以递归地添加childNodes并检查值?

MyGrammar grammar = new MyGrammar ();
Parser parser = new Parser(grammar);
var result = parser.Parse(textBox.Text);
var IfNode=result.Root.ChildNodes[0].ChildNodes[0].ChildNodes[1].ChildNodes[0].ToString() == "IfStatement";

我正在尝试这样做，但它不工作

  var IfCheck = result.Root.ChildNodes.FindAll(x => x.ChildNodes.ToString() == "IfStatement");

遍历AST树并获得子节点值Irony

你可以遍历你的树:

/// <summary>
/// Parser extension methods
/// </summary>
public static class ParserExt
{
    /// <summary>
    /// Converts parser nodes tree to flat collection
    /// </summary>
    /// <param name="item"></param>
    /// <param name="childSelector"></param>
    /// <returns></returns>
    public static IEnumerable<ParseTreeNode> Traverse(this ParseTreeNode item, Func<ParseTreeNode, IEnumerable<ParseTreeNode>> childSelector)
    {
        var stack = new Stack<ParseTreeNode>();
        stack.Push(item);
        while (stack.Any())
        {
            var next = stack.Pop();
            yield return next;
            var childs = childSelector(next).ToList();
            for (var childId = childs.Count - 1; childId >= 0; childId--)
            {
                stack.Push(childs[childId]);
            }
        }
    }
}

然后，循环通过:

var nodes = result.Root.Traverse(node => node.ChildNodes);
var ifStatements = nodes.Where(node => node.Term.Name.Equals("IfStatement"));
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