ANTLR - identifier with whitespace

问题:

i want identifiers that can contain whitespace.

grammar WhitespaceInSymbols;

premise :   ( options {greedy=false;} : 'IF' )  id=ID{
System.out.println($id.text);
};

ID  :   ('a'..'z'|'A'..'Z')+ (' '('a'..'z'|'A'..'Z')+)* 
;

WS  :   ' '+ {skip();}
;

When i test this with "IF statement analyzed" i get a MissingTokenException and the output "IF statement analyzed".
I thought, that by using greedy=false i could tell ANTLR to exit afer 'IF' and take it as a token. But instead the IF is part of the ID. Is there a way to achieve my goal? I already tried some variations of the greed=false-option, but without success.

回答1:

I thought, that by using greedy=false i could tell ANTLR to exit afer 'IF' and take it as a token.

No, the parser has nothing to say about the creation of tokens: the input is first tokenized and then the parser rules are applied on these tokens. So setting greedy=false has no effect.

You can do this (creating ID tokens with white spaces), but it will be a horrible solution with many predicates, and a few custom methods in the lexer doing manual look-aheads: you really, really don't want this! A much cleaner solution would be to introduce a id rule in your parser and let it match one or more ID tokens.

A demo:

grammar WhitespaceInSymbols;

premise
  :  IF id THEN EOF
  ;

id
  :  ID+
  ;

IF
  :  'IF'
  ;

THEN
  :  'THEN'
  ;

ID  
  :  ('a'..'z' | 'A'..'Z')+
  ;

WS  
  :  ' '+ {skip();}
  ;

would parse the input IF statement analyzed THEN into the following tree: