Antlr AST产生(可能)疯狂 [英] Antlr AST generating (possible) madness
问题描述
以下可能吗?我想反转"输入给antlr的输入，并使每个标记成为上一个标记的子代.
Is the following even possible? I want to "reverse" the input given to antlr and make each token a child of the previous one.
因此，对于输入(假设每个令牌都由'.'字符分隔):
So, for the input (Assume each token is separated by the '.' char) :
Stack.Overflow.Horse
我希望我的语法产生以下AST:
I would like my grammar to produce the following AST:
Horse
|---Overflow
|---Stack
到目前为止，我已经设法反转了节点，但是我无法使它们彼此成为子节点:
So far, I've managed to reverse the nodes, but I'm unable to make them children of each other:
function
: ID PERIOD function
-> function ID
| ID
;
ID : 'a'..'z'*
;
推荐答案
我认为没有简单的方法可以做到这一点.您可以这样制定规则:
I don't think there's an easy way to do that. You could make your rule like this:
function
: ID PERIOD function
-> ^(function ID)
| ID
;
但这只会使最后一个节点成为根节点，而所有其他节点为其子节点.例如，以下来源:
but that only makes the last node the root and all other nodes its children. For example, the following source:
a.b.c.d.e
将产生以下树:
e
/ / \ \
d c b a
我看不到简单的解决方法，因为当您第一次解析a.b.c.d.e时，a将是ID和b.c.d.e的递归调用function:
I can't see an easy fix since when you first parse a.b.c.d.e, a will be the ID and b.c.d.e the recursive call to function:
a.b.c.d.e
| +-----+
| |
| `-----> function
|
`----------> ID
导致b.c.d.e将以a作为其子级的事实.然后，当b成为ID时，它也作为子元素添加到a旁边.在您的情况下，应将a作为子项删除，然后将其添加到b的子项列表中.但是AFAIK，这在ANLTR中是不可能的(至少，不是在语法内部以一种清晰的方式).
resulting in the fact that b.c.d.e will have a as its child. When then b becomes the ID, it too is added as a child next to a. In your case, a should be removed as a child and then added to the list of b's children. But AFAIK, that is not possible in ANLTR (at least, not in a clean way inside the grammar).
编辑
好吧，作为一种变通办法，我脑子里有一些优雅的东西，但这并没有达到我的期望.因此，作为不太优雅的解决方案，您可以将last节点作为重写规则中的根节点:
Okay, as a work-around I had something elegant in mind, but that didn't work as I had hoped. So, as a less elegant solution, you could match the last node as the root in your rewrite rule:
function
: (id '.')* last=id -> ^($last)
;
，然后使用+=运算符收集List中所有可能的先前节点(children):
and then collect all possible preceding nodes (children) in a List using the += operator:
function
: (children+=id '.')* last=id -> ^($last)
;
并使用解析器中的自定义成员方法将这些children注入"到树的根中(在List中从右到左！):
and use a custom member-method in the parser to "inject" these children into the root of your tree (going from right to left in your List!):
function
: (children+=id '.')* last=id {reverse($children, (CommonTree)$last.tree);} -> ^($last)
;
一个小演示:
grammar ReverseTree;
options {
output=AST;
}
tokens {
ROOT;
}
@members {
private void reverse(List nodes, CommonTree root) {
if(nodes == null) return;
for(int i = nodes.size()-1; i >= 0; i--) {
CommonTree temp = (CommonTree)nodes.get(i);
root.addChild(temp);
root = temp;
}
}
}
parse
: function+ EOF -> ^(ROOT function+)
;
function
: (children+=id '.')* last=id {reverse($children, (CommonTree)$last.tree);} -> ^($last)
;
id
: ID
;
ID
: ('a'..'z' | 'A'..'Z')+
;
Space
: ' ' {skip();}
;
还有一个小测试班:
import org.antlr.runtime.*;
import org.antlr.runtime.tree.*;
import org.antlr.stringtemplate.*;
public class Main {
public static void main(String[] args) throws Exception {
ANTLRStringStream in = new ANTLRStringStream("a.b.c.d.e Stack.Overflow.Horse singleNode");
ReverseTreeLexer lexer = new ReverseTreeLexer(in);
CommonTokenStream tokens = new CommonTokenStream(lexer);
ReverseTreeParser parser = new ReverseTreeParser(tokens);
ReverseTreeParser.parse_return returnValue = parser.parse();
CommonTree tree = (CommonTree)returnValue.getTree();
DOTTreeGenerator gen = new DOTTreeGenerator();
StringTemplate st = gen.toDOT(tree);
System.out.println(st);
}
}
这将产生如下所示的AST:
which will produce an AST that looks like:
- (使用 http://graph.gafol.net 生成的图像)
- (image generated using http://graph.gafol.net)
对于输入字符串:
"a.b.c.d.e Stack.Overflow.Horse singleNode"
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