ANTLR4 integer parsing fail

1
votes

I'm trying to parse a file with antlr 4, I can't get why integer of more than one digit are not parsified (line 79:44 no viable alternative at input '17').

This is the entier grammar http://pastebin.com/rxktvUBi

Here is the definition of int 

fragment DIGIT : [0-9] ;
integer : DIGIT+ ;

which doesn't work at all. This version 

integer : ('0' | '1' | '2' | '3' | '4' | '5' | '6' | '7' | '8' | '9')+ ;

works only for 1 digit integers.

This is an example of line not parsified 

 struct p_77_bound_17_or: ((bound(MEK)<=17) | (bound(MEKPP)<=17))

The problem is in 

 simple_expression:

     (integer)+

Note that if I use identifier 

ID:
  ('a'..'z'|'A'..'Z'|'0'..'9'|'_')+;


identifier: ID;

instead of integer 

 simple_expression:

     identifier

that works.

Why? Any Idea?

1
parsingintegerparser-generatorantlr4

1 Answers

8
votes

Your integer rule is a parser rule, not a lexer rule. The '0', '1', etc. literals it references are implicitly turned into lexer rules which match a single digit each. You should make the following lexer rule instead:

INTEGER : '0'..'9'+;

Or in ANTLR 4, simply this:

INTEGER : [0-9]+;