Antlr解析树节点坐标？

Question

I use Antlr4 4.9.2

I have a requirement to perform multiple passes of the same parse tree at different stages of my analysis. Some of the files my application handles are very large, therefore I'd like to be able to avoid keeping the parse tree in memory, and be able to regenerate a different parse tree instance each time. So far so good.

My challenge is that I need a way to (a) compare nodes and (b) quickly access nodes that works with different instances of equivalent parse trees.

For example the following pseudo-code generates two separate instances of a parse tree that represent the same file (therefore the parse trees and their nodes are equivalent)

ParseTree parseTree1 = parse(myFile, myGrammar)
ParseTree parseTree2 = parse(myFile, myGrammar) 

Since myFile and myGrammar are the same, both parseTree1 and parseTree2 are equivalent, however are different instances and don't satisfy Objects.equals()

In ANTLR, how do I represent the coordinates C of a node in such a way that:

• C(node1) = C(node2) if the nodes are equivalent
• I can access C(parseTree1) or C(parseTree2) without having to visit the parse trees - so I can quickly position myself on the same node, for any instance of the parsetree

Answer 1

You can use ANTLR4's XPath implementation to directly access nodes in a given parse tree path. Here's how I get all query expressions in MySQL code, after parsing:

const expressions = XPath.findAll(tree, "/query/simpleStatement//queryExpression", this.parser);