[antlr-interest] Wildcard in tree grammar? - was validatingsemantic predicates

[Harald M. Müller]

Mhm ... but not for me. Also, I cannot see how *adding* another condition
would allow *more* alternatives to be parsed: I still get same result "no
viable ...".
The generated ifs also dont look too good, IMHO:

                    	else if ( (LA2_0 == 6) && (( x ||  !x )) )
                    	{
                    	    int LA2_2 = input.LA(2);
                    	    
                    	    if ( (LA2_2 == DOWN) && ( x ) )
                    	    {
                    	        alt2 = 1;
                    	    }
                    	    else if ( (LA2_2 == UP) && ( !x ) )
                    	    {
                    	        alt2 = 2;
                    	    }
                    	    else 
                    	    {
                    	        throw ...
                    	    }

This is exactly the same code as without the !x in the grammar, if one
removes the !x conditions; and they lead - in the case of x = false - to the
same branch, namely throw ...
Thx anyway!
Regards
Harald

> -----Original Message-----
> From: John B. Brodie [mailto:jbb at acm.org] 
> Sent: Friday, November 30, 2007 11:34 PM
> To: harald_m_mueller at gmx.de
> Cc: antlr-interest at antlr.org
> Subject: Re: [antlr-interest] Wildcard in tree grammar? - was 
> validatingsemantic predicates
> 
> I believe that you have to test the boolean condition on both 
> sides of the alternate.
> 
> Not sure whether or not that was Dr. Parr's intention.
> 
> But it seems to work for me.
> 
> >statement
> >    : simpleStatement
> >    |  
> >      ^('if' x=condition ( { x }? statement
> >                         |           .
> >                         )
> >       )
> >   ;
>    
> instead try:
> 
> statement
>     : simpleStatement
>     |  
>       ^('if' x=condition ( { x }? statement
>                          | {!x }? .
>                          )
>        )
>    ;
>    
> 
> 
> 
>    -jbb
>