Antlr4: How can I both hide and use Tokens in a grammar

问题

I'm parsing a script language that defines two types of statements; control statements and non control statements. Non control statements are always ended with ';', while control statements may end with ';' or EOL ('\n'). A part of the grammar looks like this:

script
    :   statement* EOF
    ;

statement
    :   control_statement
    |   no_control_statement
    ;

control_statement
    :   if_then_control_statement
    ;

if_then_control_statement
    :   IF expression THEN end_control_statment
        ( statement ) *
        ( ELSEIF expression THEN end_control_statment ( statement )* )*
        ( ELSE end_control_statment ( statement )* )?
        END IF end_control_statment
    ;

no_control_statement
    :   sleep_statement
    ;

sleep_statement
    :   SLEEP expression END_STATEMENT
    ;

end_control_statment
    :   END_STATEMENT
    |   EOL
    ;

END_STATEMENT
    :   ';'
    ;

ANY_SPACE
    :   ( LINE_SPACE | EOL )    ->  channel(HIDDEN)
    ;

EOL
    :   [\n\r]+
    ;

LINE_SPACE
    :   [ \t]+
    ;

In all other aspects of the script language, I never care about EOL so I use the normal lexer rules to hide white space.

This works fine in all cases but the cases where I need to use a EOL to find a termination of a control statement, but with the grammar above, all EOL is hidden and not used in the control statement rules.

Is there a way to change my grammar so that I can skip all EOL but the ones needed to terminate parts of my control statements?

回答1:

Found one way to handle this.

The idea is to divert EOL into one hidden channel and the other stuff I don´t want to see in another hidden channel (like spaces and comments). Then I use some code to backtrack the tokens when an EOL is supposed to show up and examine the previous tokens channels (since they already have been consumed). If I find something on EOL channel before I run into something from the ordinary channel, then it is ok.

It looks like this:

Changed the lexer rules:

@lexer::members {
    public static int EOL_CHANNEL = 1;
    public static int OTHER_CHANNEL = 2;
}

...

EOL
  : '\r'? '\n'  ->  channel(EOL_CHANNEL)
  ;

LINE_SPACE
  : [ \t]+  ->  channel(OTHER_CHANNEL)
  ;

I also diverted all other HIDDEN channels (comments) to the OTHER_CHANNEL. Then I changed the rule end_control_statment:

end_control_statment
  : END_STATEMENT
  | { isEOLPrevious() }?
  ;

and added

@parser::members {
  public static int EOL_CHANNEL = 1;
  public static int OTHER_CHANNEL = 2;

  boolean isEOLPrevious()
  {
        int idx = getCurrentToken().getTokenIndex();
        int ch;

        do
        {
            ch = getTokenStream().get(--idx).getChannel();
        }
        while (ch == OTHER_CHANNEL);

        // Channel 1 is only carrying EOL, no need to check token itself
        return (ch == EOL_CHANNEL);
     }
}

One could stick to the ordinary hidden channel but then there is a need to both track channel and tokens while backtracking so this is maybe a bit easier...

Hope this could help someone else dealing with these kind of issues...

来源：https://stackoverflow.com/questions/41667217/antlr4-how-can-i-both-hide-and-use-tokens-in-a-grammar

标签

parsing

whitespace

antlr4