[antlr-interest] Some bugs (or features?) in Honey Badger

[Kyle Ferrio]

I think I did not run into this because I prefer that assignments are not
expressions in my DSLs.  That said, your example reminds me of resolving
ambiguities with bison and I wonder if there's something there which could
help honey badger...

Kyle
On Feb 20, 2012 5:42 AM, "Jan Finis" <finis at in.tum.de> wrote:

> On 02/19/2012 10:33 PM, Terence Parr wrote:
> >
> > Hi. this suprising me. It translates to:
> >
> > expr[int _p]
> >      :   ( ID '=' expr[3]
> >          | ID
> >          )
> >          ( {1>= $_p}? '+' expr[2]
> >          )*
> >      ;
> >
> > (See -Xlog option).  Pretty hard for that to match as a=(a+a). are you
> sure?
>
> Hi Ter,
>
> I tested it again and was able to confirm the precedence bug, here is an
> example grammar producing the bug:
>
> grammar TestGrammar;
>
> start returns [String result]
>   : expr {$result = $expr.result; }
>   ;
>
> expr returns [String result]
>     :   ID '=' e1=expr { $result = "(" + $ID.getText() + "=" +
> $e1.result + ")"; }
>     |   ID { $result = $ID.getText(); }
>     |   e1=expr '+' e2=expr { $result = "(" + $e1.result + "+" +
> $e2.result + ")"; }
>     ;
>
> ID  :    ('a'..'z'|'A'..'Z'|'_') ('a'..'z'|'A'..'Z'|'0'..'9'|'_')*
>     ;
>
> This is the input:
>
> a=a+a
>
> The output is (a=(a+a)). With correct precedence the output should be
> ((a=a)+a).
>
> I used the jar from
>
> http://antlr.org/download/antlr-4.0ea-complete.jar
>
> and redownloaded it to make sure that I do not have an outdated version.
>
> The output was produced using this code:
>
> TestGrammarLexer lex = new TestGrammarLexer(new ANTLRInputStream(new
> FileInputStream(new File("test.input"))));
> CommonTokenStream tokens = new CommonTokenStream(lex);
>
> StartContext i = new TestGrammarParser(tokens).start();
>
> System.out.println(i.result);
>
> >
> >> the precedence should be from top to bottom, right? So, the input  a=a+a
> >> should be parsed as (a=a)+a, since the assignment rule is on the top.
> >> However, this is not the case, instead, it is parsed as a=(a+a). Bug, or
> >> am I interpreting something wrong?
> >>
> >> 2. Name binding
> >>
> >> Consider this example:
> >>
> >> expr returns [int r]
> >>      : '-' expr { $r = - $expr.r; }
> >>
> >> In this example $expr should bind to the sub-expression in my opinion.
> >> However, it does not. Since the rule is also named expr, $expr refers to
> >> the rule context instead of the context of the sub-expression. I think
> >> most of the time this is not what the user wants.
> > I think this is consistent with v3. i'll add to list to think about.
> thanks!
> > Ter
>
> Yes, it is consistent with v3, however v3 didn't have these crazy left
> recursive rules :).
> With these rules, it is much more common to have a non-terminal of the
> same type as the rule itself.
>
>
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