simple math expression parser - antlr

Question

I have a simple math expression parser and I want to build the AST by myself (means no ast parser). But every node can just hold two operands. So a 2+3+4 will result in a tree like this:
+
/ \
2 +
/ \
3 4
The problem is, that I am not able to get my grammer doing the recursion, here ist just the "add" part:
add returns [Expression e]
: op1=multiply { $e = $op1.e; Print.ln($op1.text); }
( '+' op2=multiply { $e = new AddOperator($op1.e, $op2.e); Print.ln($op1.e.getClass(), $op1.text, "+", $op2.e.getClass(), $op2.text); }
| '-' op2=multiply { $e = null; } // new MinusOperator
)*
;
But at the end of the day this will produce a single tree like:
+
/ \
2 4
I know where the problem is, it is because a "add" can occour never or infinitly (*) but I do not know how to solve this. I thought of something like:
"add" part:
add returns [Expression e]
: op1=multiply { $e = $op1.e; Print.ln($op1.text); }
( '+' op2=(multiply|add) { $e = new AddOperator($op1.e, $op2.e); Print.ln($op1.e.getClass(), $op1.text, "+", $op2.e.getClass(), $op2.text); }
| '-' op2=multiply { $e = null; } // new MinusOperator
)?
;
But this will give me a recoursion error. Any ideas?

Answer 1

I don't have the full grammar to test this solution, but consider replacing this (from the first add rule in the question):
$e = new AddOperator($op1.e, $op2.e);
With this:
$e = new AddOperator($e, $op2.e); //$e instead of $op1.e
This way each iteration over ('+' multiply)* extends e rather than replaces it.
It may require a little playing around to get it right, or you may need a temporary Expression in the rule to keep things managed. Just make sure that the last expression created by the loop is somewhere on the right-hand side of the = operator, as in $e = new XYZ($e, $rhs.e);.