Parse stacked comparison expression into logical Conjunction Tree with antlr3 - parsing

Question

I have run into a problem, when i tried to parse a stacked arithmetic comparison expression:
"1<2<3<4<5"
into a logical Tree of Conjunctions:
CONJUNCTION(COMPARISON(1,2,<) COMPARISON(2,3,<) COMPARISON(3,4,<) COMPARISON(4,5,<))
Is there a way in Antlr3 Tree Rewrite rules to iterate through matched tokens and create the result Tree from them in the target language (I'm using java)? So i could make COMPARISON nodes from element x, x-1 of matched 'addition' tokens. I know i can reference the last result of a rule but that way i'd only get nested COMPARISON rules, that's not what i wish for.
/This is how i approached the problem, sadly it doesn't do what i would like to do yet of course.
fragment COMPARISON:;
operator
:
('<'|'>'|'<='|'>='|'=='|'!=')
;
comparison
#init{boolean secondpart = false;}
:
e=addition (operator {secondpart=true;} k=addition)*
-> {secondpart}? ^(COMPARISON ^(VALUES addition*) ^(OPERATORS operator*))
-> $e
;
//Right now what this does is:
tree=(COMPARISON (VALUES (INTEGERVALUE (VALUE 1)) (INTEGERVALUE (VALUE 2)) (INTEGERVALUE (VALUE 3)) (INTEGERVALUE (VALUE 4)) (INTEGERVALUE (VALUE 5))) (OPERATORS < < < <))
//The label for the CONJUNCTION TreeNode that i would like to use:
fragment CONJUNCTION:;

Answer 1

I came up with a nasty solution to this problem by writing actual tree building java code:
grammar testgrammarforcomparison;
options {
language = Java;
output = AST;
}
tokens
{
CONJUNCTION;
COMPARISON;
OPERATOR;
ADDITION;
}
WS
:
('\t' | '\f' | ' ' | '\r' | '\n' )+
{$channel = HIDDEN;}
;
comparison
#init
{
List<Object> additions = new ArrayList<Object>();
List<Object> operators = new ArrayList<Object>();
boolean secondpart = false;
}
:
(( e=addition {additions.add(e.getTree());} ) ( op=operator k=addition {additions.add(k.getTree()); operators.add(op.getTree()); secondpart = true;} )*)
{
if(secondpart)
{
root_0 = (Object)adaptor.nil();
Object root_1 = (Object)adaptor.nil();
root_1 = (Object)adaptor.becomeRoot(
(Object)adaptor.create(CONJUNCTION, "CONJUNCTION")
, root_1);
Object lastaddition = additions.get(0);
for(int i=1;i<additions.size();i++)
{
Object root_2 = (Object)adaptor.nil();
root_2 = (Object)adaptor.becomeRoot(
(Object)adaptor.create(COMPARISON, "COMPARISON")
, root_2);
adaptor.addChild(root_2, additions.get(i-1));
adaptor.addChild(root_2, operators.get(i-1));
adaptor.addChild(root_2, additions.get(i));
adaptor.addChild(root_1, root_2);
}
adaptor.addChild(root_0, root_1);
}
else
{
root_0 = (Object)adaptor.nil();
adaptor.addChild(root_0, e.getTree());
}
}
;
/** lowercase letters */
fragment LOWCHAR
: 'a'..'z';
/** uppercase letters */
fragment HIGHCHAR
: 'A'..'Z';
/** numbers */
fragment DIGIT
: '0'..'9';
fragment LETTER
: LOWCHAR
| HIGHCHAR
;
IDENTIFIER
:
LETTER (LETTER|DIGIT)*
;
addition
:
IDENTIFIER ->^(ADDITION IDENTIFIER)
;
operator
:
('<'|'>') ->^(OPERATOR '<'* '>'*)
;
parse
:
comparison EOF
;
For input
"DATA1 < DATA2 > DATA3"
This outputs tree such as:
If you guys know any better solutions, please tell me about them