词法分析时，ANTLR可以返回代码行吗?

问题描述

我正在尝试使用ANTLR使用完整的Java语法分析大量代码.由于ANTLR需要打开所有源文件并对其进行扫描，所以我想知道它是否还可以返回代码行.

I am trying use ANTLR to analyse a large set of code using full Java grammar. Since ANTLR needs to open all the source files and scan them, I am wondering if it can also return lines of code.

我检查了Lexer和Parser的API，看来它们没有返回LoC.一点点地掌握语法规则来获得LoC容易吗?完整的Java规则很复杂，我真的不想弄乱它的很大一部分.

I checked API for Lexer and Parser, it seems they do not return LoC. Is it easy to instrument the grammar rule a bit to get LoC? The full Java rule is complicated, I don't really want to mess a large part of it.

推荐答案

如果您已有ANTLR语法，并且想在解析过程中计算某些内容，则可以执行以下操作:

If you have an existing ANTLR grammar, and want to count certain things during parsing, you could do something like this:

grammar ExistingGrammar;

// ...

@parser::members {
  public int loc = 0;
}

// ...

someParserRule
 : SomeLexerRule someOtherParserRule {loc++;}
 ;

// ...

因此，每当操作者遇到someParserRule时，都可以通过将{loc++;}放置在规则之后(或之前)来将loc增加1.

So, whenever your oparser encounters a someParserRule, you increase the loc by one by placing {loc++;} after (or before) the rule.

因此，无论您对代码行的定义是什么，只需将{loc++;}放在规则中以增加计数器.注意不要增加两次:

So, whatever your definition of a line of code is, simply place {loc++;} in the rule to increase the counter. Be careful not to increase it twice:

statement
 : someParserRule {loc++;}
 | // ...
 ;

someParserRule
 : SomeLexerRule someOtherParserRule {loc++;}
 ;

编辑

我刚刚注意到，在您的问题的标题中，您询问是否可以在词汇化过程中完成此操作.那是不可能的.假设LoC总是以';'结尾.在词法分析过程中，您将无法在例如赋值后的';'和单个for(int i = 0; i < n; i++) { ... }语句中的两个';'之间进行区分，例如，赋值是单个LoC.设为2 LoC.

EDIT

I just noticed that in the title of your question you asked if this can be done during lexing. That won't be possible. Let's say a LoC would always end with a ';'. During lexing, you wouldn't be able to make a distinction between a ';' after, say, an assignment (which is a single LoC), and the 2 ';'s inside a for(int i = 0; i < n; i++) { ... } statement (which wouldn't be 2 LoC).

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