Resolving ANTLR ambiguity while matching specific Types

Question

I'm starting exploring ANTLR and I'm trying to match this format: (test123 A0020 )

Where :

• test123 is an Identifier of max 10 characters ( letters and digits )
• A : Time indicator ( for Am or Pm ), one letter can be either "A" or "P"
• 0020 : 4 digit format representing the time.

I tried this grammar :

    IDENTIFIER
:
    ( LETTER | DIGIT ) +
;
    INT
:
    DIGIT+
;
fragment
DIGIT
:
    [0-9]
;

fragment
LETTER
:
    [A-Z]
;

WS : [ \t\r\n(\s)+]+ -> channel(HIDDEN) ;
formatter:  '(' information ')';

information : 
information '/' 'A' INT 
        |IDENTIFIER ;

How can I resolve the ambiguity and get the time format matched as 'A' INT not as IDENTIFIER? Also how can I add checks like length of token to the identifier? I tknow that this doesn't work in ANTLR : IDENTIFIER : (DIGIT | LETTER ) {2,10}

UPDATE:

I changed the rules to have semantic checks but I still have the same ambiguity between the identifier and the Time format. here's the modified rules:

formatter
    : information
    | information '-' time
    ;

time :
    timeMode timeCode;  

timeMode:   
    { getCurrentToken().getText().matches("[A,C]")}? MOD
;

timeCode: {getCurrentToken().getText().matches("[0-9]{4}")}?  INT;

information: {getCurrentToken().getText().length() <= 10 }? IDENTIFIER;

MOD:  'A' | 'C';

So the problem is illustrated in the production tree, A0023 is matched to timeMode and the parser is complaining that the timeCode is missing

Answer 1

Here is a way to handle it:

grammar Test;

@lexer::members {
  private boolean isAhead(int maxAmountOfCharacters, String pattern) {
    final Interval ahead = new Interval(this._tokenStartCharIndex, this._tokenStartCharIndex + maxAmountOfCharacters - 1);
    return this._input.getText(ahead).matches(pattern);
  }
}

parse
 : formatter EOF
 ;

formatter
 : information ( '-' time )?
 ;

time
 : timeMode timeCode
 ;

timeMode
 : TIME_MODE
 ;

timeCode
 : {getCurrentToken().getType() == IDENTIFIER_OR_INTEGER && getCurrentToken().getText().matches("\\d{4}")}?
   IDENTIFIER_OR_INTEGER
 ;

information
 : {getCurrentToken().getType() == IDENTIFIER_OR_INTEGER && getCurrentToken().getText().matches("\\w*[a-zA-Z]\\w*")}?
   IDENTIFIER_OR_INTEGER
 ;

IDENTIFIER_OR_INTEGER
 : {!isAhead(6, "[AP]\\d{4}(\\D|$)")}? [a-zA-Z0-9]+
 ;

TIME_MODE
 : [AP]
 ;

SPACES
 : [ \t\r\n] -> skip
 ;

A small test class:

public class Main {

    private static void indent(String lispTree) {

        int indentation = -1;

        for (final char c : lispTree.toCharArray()) {
            if (c == '(') {
                indentation++;
                for (int i = 0; i < indentation; i++) {
                    System.out.print(i == 0 ? "\n  " : "  ");
                }
            }
            else if (c == ')') {
                indentation--;
            }
            System.out.print(c);
        }
    }

    public static void main(String[] args) throws Exception {
        TestLexer lexer = new TestLexer(new ANTLRInputStream("1P23 - A0023"));
        TestParser parser = new TestParser(new CommonTokenStream(lexer));
        indent(parser.parse().toStringTree(parser));
    }
}

will print:

(parse 
  (formatter 
    (information 1P23) - 
    (time 
      (timeMode A) 
      (timeCode 0023))) <EOF>)

for the input "1P23 - A0023".

EDIT

ANTLR also can output the parse tree on UI component. If you do this instead:

public class Main {

    public static void main(String[] args) throws Exception {
        TestLexer lexer = new TestLexer(new ANTLRInputStream("1P23 - A0023"));
        TestParser parser = new TestParser(new CommonTokenStream(lexer));
        new TreeViewer(Arrays.asList(TestParser.ruleNames), parser.parse()).open();
    }
}

the following dialog will appear:

Tested with ANTLR version 4.5.2-1