Antlr 3.3 return values in java

Question

I try to figure out how to get values from the parser. My input is 'play the who' and it should return a string with 'the who'.

Sample.g:

text returns [String value]
 : speech = wordExp space name {$value = $speech.text;}
 ;

name returns [String value] 
 : SongArtist = WORD (space WORD)*  {$value = $SongArtist.text;}
 ;

wordExp returns [String value] 
 : command = PLAY {$value = $command.text;} | command = SEARCH {$value = $command.text;}
 ; 

PLAY   : 'play';
SEARCH : 'search';
space : ' ';

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


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

If I enter 'play the who' that tree comes up:

http://i.stack.imgur.com/ET61P.png

I created a Java file to catch the output. If I call parser.wordExp() I supposed to get 'the who', but it returns the object and this EOF failure (see the output below). parser.text() returns 'play'.

import org.antlr.runtime.*;

import a.b.c.SampleLexer;
import a.b.c.SampleParser;

public class Main {
    public static void main(String[] args) throws Exception {
        ANTLRStringStream in = new ANTLRStringStream("play the who");

        SampleLexer lexer = new SampleLexer(in);
        CommonTokenStream tokens = new CommonTokenStream(lexer);
        SampleParser parser = new SampleParser(tokens);

        System.out.println(parser.text());
        System.out.println(parser.wordExp());
    }
}

The console return this:

play
a.b.c.SampleParser$wordExp_return@1d0ca25a
line 1:12 no viable alternative at input '<EOF>'

How can I catch 'the who'? It is weird for me why I can not catch this string. The interpreter creates the tree correctly.

Answer 1

First, in your grammar, speech only gets assigned the return value of parser rule wordExp . If you want to manipulate the return value of rule name as well, you can do this with an additional variable like the example below.

text returns [String value]
 : a=wordExp space b=name {$value = $a.text+" "+$b.text;}
 ;

Second, invoking parser.text() parses the entire input. A second invocation (in your case parser.wordExp() ) thus finds EOF. If you remove the second call the no viable alternative at input 'EOF' goes away.

There may be a better way to do this, but in the meantime this may help you out.