Parse comment line

Question

Given the following basic grammar I want to understand how I can handle comment lines. Missing is the handling of the <CR><LF> which usually terminates the comment line - the only exception is a last comment line before the EOF, e. g.:

# comment
abcd := 12 ;
# comment eof without <CR><LF>

grammar CommentLine1a;

//==========================================================
// Options
//==========================================================



//==========================================================
// Lexer Rules
//==========================================================

Int
  : Digit+
  ;

fragment Digit
  : '0'..'9'
  ;

ID_NoDigitStart
  : ( 'a'..'z' | 'A'..'Z' ) ('a'..'z' | 'A'..'Z' | Digit )*
  ;

Whitespace
  : ( ' ' | '\t' | '\r' | '\n' )+ { $channel = HIDDEN ; }
  ; 


//==========================================================
// Parser Rules
//==========================================================

code
  : ( assignment | comment )+
  ;

assignment
  : id_NoDigitStart ':=' id_DigitStart ';'
  ;

id_NoDigitStart
  : ID_NoDigitStart
  ;  

id_DigitStart
  : ( ID_NoDigitStart | Int )+
  ;

comment
  : '#' ~( '\r' | '\n' )*
  ;

Answer 1

Unless you have a very compelling reason to put the comment inside the parser (which I'd like to hear), you should put it in the lexer:

Comment
  :  '#' ~( '\r' | '\n' )*
  ;

And since you already account for line breaks in your Space rule, there's no problem with input like # comment eof without <CR><LF>

Also, if you use literal tokens inside parser rules, ANTLR automatically creates lexer rules of them behind the scenes. So in your case:

comment
  :  '#' ~( '\r' | '\n' )*
  ;

would match a '#' followed by zero or more tokens other than '\r' and '\n' and not zero or more characters other than '\r' and '\n'.

For future reference:

Inside parser rules

• ~ negates tokens
• . matches any token

Inside lexer rules

• ~ negates characters
• . matches any character in the range 0x0000 ... 0xFFFF