I'm rewriting a tree in a way that I think follows the rules. Here's
a sample input fragment:
"one" == "two" && "three" == "four" && "five" == "six"
The following rewrite works correctly, but then I have to iterate over
the list and match rhs-lhs pairs,a mere minor annoyance I suppose:
matchExpression
: (object EQUALS object) (AND object EQUALS object)*
-> ^(AST_MATCH object+)
;
This version gives the following tree:
(AST_MATCH "one" "two" "three" "four" "five" "six")
I'd prefer to have a list of the right-hand side and one of the
left-hand side, so I tried the following:
matchExpression
: (rhs=object EQUALS lhs=object) (AND rhs=object EQUALS lhs=object)*
-> ^(AST_MATCH ^(AST_MATCH_ARG $rhs+) ^(AST_MATCH_ARG $lhs+))
;
The above yields the following tree:
(AST_MATCH (AST_MATCH_ARG "five") (AST_MATCH_ARG "six"))
But I expected the following:
(AST_MATCH (AST_MATCH_ARG "one" "three" "five") (AST_MATCH_ARG "two"
"four" "six"))
As 'rhs' and 'lhs' are indeed present more than once, and being
matched more than once, shouldn't the rewritten tree contain all the
nodes or am I misunderstanding something?
Thanks.
--
Kaleb Pederson
Blog - http://kalebpederson.com
Twitter - http://twitter.com/kalebpederson
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