[antlr-interest] parse tree construction

[John B. Brodie]

Greetings!

Martijn's suggestions are close but need a couple of tweaks, see
below...

On Tue, 2009-08-25 at 18:50 +0200, Martijn Reuvers wrote:
> Hi Safiye,
> 
> As Sam said, recursion might be what you need.
> 
> Something like this gives your tree I believe:
> 
> grammar test;
> 
> options {
> 	output=AST;
> 	
> }
> 
> x
> 	: A B y    -> ^(B A y)
> 	;

this will REQUIRE at least 1 instance of the C A B clause. whereas the
original rule had it being 0 or more. to fix make the y be y?, e.g:

x : A B y? -> ^(B A y?) ;

> y :	C A B y* -> ^(B A C y*)
> 	;

this is a recursive application of the y rule inside a loop! i suspect
it will work but seems very dangerous to me. i would suggest using the ?
operator rather than the * operator, 0 or 1 vs 0 or many; e.g:

y : C A B y? -> ^(B A C y?) ;

> 
> A :	'A';
> 
> B :	'B';
> 
> C :	'C';
> 
> 
> Martijn
> 
> On Tue, Aug 25, 2009 at 4:57 PM, Safiye Celik<safisce at gmail.com> wrote:
> > Hi
> > I wanna ask one more question about parse tree construction..
> > I have such a rule:
> >
> > x: A B (C A B)* and I have to create such a tree for it:
> >
> >
> >                                            B
> >                                      /      |     \
> >                                 A        C        B
> >                                                   /  |  \
> >                                                A  C    B
> >                                                         /  |  \
> >                                                       A..C. B......and
> > continues like that. (The leftmost a B is in the rule, the closest it is to
> > the root of the tree.This is true also for other tokens)
> >
> > How can I create such a tree for that rule? the "*" confuses everything :(
> >
> >
> > --
> > -safiye

Hope this helps...
   -jbb