Antlr parser doesn't generate all expected code

Question

I have a grammar as roughly defined in Tutorial for walking ANTLR ASTs in C#?:

grammar Test; 

options 
{
language = 'CSharp3'; 
output=AST; 
} 
public expr : mexpr (PLUS^ mexpr)* SEMI! 
; 
mexpr 
: atom (STAR^ atom)* 
; 
atom: INT 
; 
//class csharpTestLexer extends Lexer; 
WS : (' ' 
| '\t' 
| '\n' 
| '\r') 
{ $channel = Hidden; } 
; 
LPAREN: '(' 
; 
RPAREN: ')' 
; 
STAR: '*' 
; 
PLUS: '+' 
; 
SEMI: ';' 
; 
protected 
DIGIT 
: '0'..'9' 
; 
INT : (DIGIT)+ 
;

This builds, but leaves me with no parser.expr_result class which I did expect, and parser.expr() returns AstParserRuleReturnScope what am I doing wrong? Are it the options? tool commandline options? anything else?

user1201210 · Accepted Answer · 2012-12-04 17:11:37Z

1

ANTLR 3.3 declares rule expr like so:

    public TestParser.expr_return expr()

ANTLR 3.4 declares it like so:

    public AstParserRuleReturnScope<object, IToken> expr()

Here's the definition of TestParser.expr_return:

    public class expr_return : ParserRuleReturnScope<IToken>, IAstRuleReturnScope<object>
    {
        private object _tree;
        public object Tree { get { return _tree; } set { _tree = value; } }
    }

The AstParserRuleReturnScope class appears equivalent to the generated expr_return class.

answered Dec 4, 2012 at 17:11

user1201210

3,8091 gold badge25 silver badges26 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Antlr parser doesn't generate all expected code

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related