[antlr-interest] AST Tree generation question

[Gavin Lambert]

At 21:26 12/12/2007, Roberto Attias wrote:
 >In an expression grammar derived from one of the examples I have 

 >the following rule:
 >
 >additive_expression
 >    : (multiplicative_expression) ('+^' 
multiplicative_expression |
 >'-'^
 >multiplicative_expression)*
 >    ;
 >
 >When parsing a+b", this generates a tree as expected. Now ,I 
want
 >to have OP_ADD or OP_SUB as radix of the tree when evaluating 
a+b
 >or a-b.

The easiest way to do this would be to declare OP_ADD as a token 
consisting of '+', and use that in place of the constant 
above.  (And that ^ should be outside the quotes; I'm assuming 
that's a typo.)

If you don't want to do that for some reason, then you can use 
rewrites.  I'm actually not completely sure of the exact syntax 
myself, but I think you can do something like this:

additive_expression
   : (a=multiplicative_expression -> $a)
     (
       (b='+' c=multiplicative_expression
          -> ^(OP_ADD[$b] $additive_expression $c))
     | (d='-' e=multiplicative_expression
          -> ^(OP_SUB[$d] $additive_expression $e))
     )*
   ;

The key points are that a rewrite arrow sets the tree output for 
the entire rule, no matter where it appears, and once set you can 
refer to the previously-set tree when building another tree.  The 
other useful bit is the square brackets after the OP_* virtual 
tokens; those aren't required, but if you use them it'll associate 
the corresponding source token (the operator) with the tree node, 
so it'll still have text and a source location.