Question

I already know the workaround for this problem, but I would like to really use this one approach, for at least one reason — it should work.

This is rule taken from “The Definitive ANTLR Reference” by Terence Parr (the books is for ANTLR3):

expr : (INT -> INT) ('+' i=INT -> ^('+' $expr $i) )*;

If INT is not followed by + the result will be INT (single node), if it is — subtree will be built with first INT (referred as $expr) as left branch.

I would like to build similar rule, yet with custom action:

mult_expr : (pow_expr -> pow_expr ) 
            (op=MUL exr=pow_expr 
              -> { new BinExpr($op,$mult_expr.tree,$exr.tree) })*; 

ANTLR accepts such rule, but when I run my parser with input (for example) “5 * 3” it gives me an error “line 1:1 missing EOF at ‘*’5”.

QUESTION: how to use back reference with custom rewrite action?

Answer 1

I am persistent guy, and this idea of using my custom nodes in one step was bothering me… 😉

So, I did. The crucial points are:

• putting EOF! at the end of the “main” rule,

• when labeling the tokens, putting labels next to token, not to group, so (op='*'|op='/'), not op=('*'|'/')

I don’t know for sure if this approach of using grammar rules to create immediately custom nodes will be a good a idea, but since this solves the problem asked in question I am marking this as solution.

And for the record, the most interesting rule looks now like this:

mult_expr : (exl=pow_expr -> $exl ) 
        ((op=MUL|op=IDIV|op=RDIV|op=MOD) exr=pow_expr 
        -> { new BinaryExpression($op,$exl.tree,$exr.tree) })*;