ANTLR 复制一棵树 [英] ANTLR duplicate a tree

问题描述

我使用ANTLR构建了一棵树(CommonTree)，就像follwing(语言:JAVA):

 Parser.prog_return r = parser.prog();CommonTree t = (CommonTree) r.getTree();

现在，我需要将t"作为参数传递，并在不影响原始树的情况下进行一些更改.但是，使用 Java 的指针，这无法完成，因此我需要复制树.

我在网上搜索过，我能找到的最隐蔽的是ASTFactory类的dupTree()方法.

关于如何实现这一点的任何建议或建议将不胜感激！

编辑

@Bart Kiers，感谢您的回答，它绝对有效！

我看到您正在对树进行深度优先遍历，并为每个访问过的节点创建一个 CommonTree 对象.

我现在的问题是，CommonToken 和 CommonTree 之间的关系是什么，这些属性有什么作用:

cTok.setCharPositionInLine(oTok.getCharPositionInLine());cTok.setChannel(oTok.getChannel());cTok.setStartIndex(oTok.getStartIndex());cTok.setStopIndex(oTok.getStopIndex());cTok.setTokenIndex(oTok.getTokenIndex());

解决方案

试试这个:

public static CommonTree copyTree(CommonTree original) {CommonTree copy = new CommonTree(original.getToken());copyTreeRecursive(copy, original);返回副本；}private static void copyTreeRecursive(CommonTree copy, CommonTree original) {if(original.getChildren() != null) {for(Object o : original.getChildren()) {CommonTree originalChild = (CommonTree)o;//从原始子节点获取tokenCommonToken oTok = (CommonToken)originalChild.getToken();//创建一个具有相同类型的新令牌 &文本为oTok"CommonToken cTok = new CommonToken(oTok.getType(), oTok.getText());//将所有属性从 'oTok' 复制到 'cTok'cTok.setLine(oTok.getLine());cTok.setCharPositionInLine(oTok.getCharPositionInLine());cTok.setChannel(oTok.getChannel());cTok.setStartIndex(oTok.getStartIndex());cTok.setStopIndex(oTok.getStopIndex());cTok.setTokenIndex(oTok.getTokenIndex());//创建一个新的树节点，以 'cTok' 作为标记CommonTree copyChild = new CommonTree(cTok);//设置子节点的父节点copyChild.setParent(copy);//将子节点添加到父节点copy.addChild(copyChild);//进行递归调用以进行更深的复制copyTreeRecursive(copyChild, originalChild);}}}...//获取原始树CommonTree 树 = (CommonTree)parser.parse().getTree();//创建树的副本CommonTree copy = copyTree(tree);//改变根的右节点的右节点的内容((CommonTree)tree.getChild(1).getChild(1)).getToken().setText("X");System.out.println(tree.toStringTree());System.out.println(copy.toStringTree());

会产生:

<前>(&& a (|| b X))(&& a (|| b c))

对于输入a && (b || c)".即，tree 有 X，但 copy 将具有原始内容:c.

请注意，我选择了 CommonTree 和 CommonToken 对象，因为它们是默认的 Token 和 Tree 实现.如果您选择创建自己的 Token 和/或 Tree，您可能会继承 CommonTree 和 CommonToken 类，在这种情况下，我的建议不会中断.

一个CommonTree 只不过是一个CommonToken 的包装器，包含一些额外的信息:父节点和子节点.这就是为什么我还要从 CommonToken 对象中复制所有信息.

I use ANTLR to build a tree (CommonTree) like follwing (language: JAVA):

  Parser.prog_return r = parser.prog();
  CommonTree t = (CommonTree) r.getTree();

Now, I need to pass "t" as a parameter, and make some changes without affecting the original tree. However, with Java's pointer, this could not been done, so I need to duplicate the tree.

I have been search on the internet, the cloested thing I could find is the dupTree() method of ASTFactory class.

Any suggestion or advises on how to achive this would be appreciated!

EDIT

@Bart Kiers, Thanks for your answer, it absolutely works!

I see you are doing a depth first walk over the tree, and create a CommonTree object for each node that was visited.

My question is now, what is the relation between CommonToken and CommonTree, and what are these attributes do:

cTok.setCharPositionInLine(oTok.getCharPositionInLine());
cTok.setChannel(oTok.getChannel());
cTok.setStartIndex(oTok.getStartIndex());
cTok.setStopIndex(oTok.getStopIndex());
cTok.setTokenIndex(oTok.getTokenIndex());

解决方案

Try something like this:

public static CommonTree copyTree(CommonTree original) {
  CommonTree copy = new CommonTree(original.getToken());
  copyTreeRecursive(copy, original);
  return copy;
}

private static void copyTreeRecursive(CommonTree copy, CommonTree original) {
  if(original.getChildren() != null) {
    for(Object o : original.getChildren()) {

      CommonTree originalChild = (CommonTree)o;

      // get the token from the original child node
      CommonToken oTok = (CommonToken)originalChild.getToken();

      // create a new token with the same type & text as 'oTok' 
      CommonToken cTok = new CommonToken(oTok.getType(), oTok.getText());

      // copy all attributes from 'oTok' to 'cTok'  
      cTok.setLine(oTok.getLine());
      cTok.setCharPositionInLine(oTok.getCharPositionInLine());
      cTok.setChannel(oTok.getChannel());
      cTok.setStartIndex(oTok.getStartIndex());
      cTok.setStopIndex(oTok.getStopIndex());
      cTok.setTokenIndex(oTok.getTokenIndex());

      // create a new tree node with the 'cTok' as token
      CommonTree copyChild = new CommonTree(cTok);

      // set the parent node of the child node
      copyChild.setParent(copy);

      // add the child to the parent node
      copy.addChild(copyChild);

      // make a recursive call to copy deeper
      copyTreeRecursive(copyChild, originalChild);
    }
  }
}

...

// get the original tree
CommonTree tree = (CommonTree)parser.parse().getTree();

// create a copy of the tree
CommonTree copy = copyTree(tree);

// change the contents of the right node of the right node of the root 
((CommonTree)tree.getChild(1).getChild(1)).getToken().setText("X");

System.out.println(tree.toStringTree());
System.out.println(copy.toStringTree());

which would produce:

(&& a (|| b X))
(&& a (|| b c))

for the input "a && (b || c)". I.e., the tree has X, but the copy would will have the original contents: c.

Note that I choose CommonTree and CommonToken objects because those are the default Token and Tree implementations. If you choose to create your own Token and/or Tree, chances are that you'll subclass the CommonTree and CommonToken classes, in which case my suggestion would not break.

A CommonTree is nothing more than a wrapper around a CommonToken, holding a bit of extra information: a parent node, and child nodes. That is why I also copy all the information from the CommonToken objects.

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